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THE 
STRENGTH OF MATERIALS 



BY THE SAME AUTHOR 



THE THEORY AND DESIGN 
OF STRUCTURES 

Third Edition. 618 pp. Demy 8vo. 95. net. 

FURTHER PROBLEMS IN THE 
THEORY AND DESIGN OF 
STRUCTURES 

236 pp. Demy 8vo. 75. 6f/. net. 

ALIGNMENT CHARTS 

32 pp. Crown 8vo. \s. 2,d. net. 



London: Chapman and Hall, Ltd., 11 Henrietta Street, W'.C. 

THE ELEMENTARY PRINCIPLES 
OF REINFORCED CONCRETE 
CONSTRUCTION 

200 pp. Crown 8vo. 3^. net. 

CALCULUS FOR ENGINEERS 

(With Dr. H. BRYON HEYWOOD). 
269 pp. Crown 8vo. ^s. net. 
[/// the Broadway Scries of Engineerhtg Handbooks.^ 



London : Scott, Greenwood & Son, 8 Broadway, 
Ludgate Hill, E.G. 



AN INTRODUCTION TO 
APPLIED MECHANICS 

309 pp. Demy 8vo. 4J-. dd. net. 

\In the Cambridge Technical Series\ 



Cambridge : at the University Press. 



THE STRENGTH OF 
MATERIALS 

A TEXT-BOOK FOR 
ENGINEERS AND ARCHITECTS 



y 



BY .'i^' 

EWART S.^^'aNDREWS 

B.Sc, Eng. (Lond.) 

MEMBEk OF THE CONCRETE INSTITUTE; LECTUKEK IN THE ENGINEEUING 

DEPARTMENT OF THE GOLDSMITHS' COLLEtJK, NEW CROSS, AND 

AT THE WESTMINSTER TECHNICAL INSTITUTE ; FORMERLY 

DEMONSTRATOR AND LECTURER IN THE ENGINEERING 

DEPARTMENT OF UNIVERSITY COLLEGE, 

LONDON 

AUTHOR OF "theory AND DliSIGN OF STRUCTURES," "REINFORCED 
CONCRETE CONSTRUCTION," "CALCULUS FOR ENGINEERS," ETC. 



fVlTH NUMEROUS ILLUSTRATIONS, TABLES 
AND WORKED EXAMPLES 



NEW YORK 
D. VAN NOSTRAND COMPANY 

TWENTY-FIVE PARK PLACE 

1916 

\All rights reserved '\ 






Printed in Great Britain by 

Richard Clay & Sons, Limited, 

brunswick st., stamford st., s.e., 

and bungay suffolk. 



l^ 



It 



^ 



PREFACE 

The advance in the application of scientific methods to 
architectural and engineering problems has made increasing 
demands upon the theoretical knowledge required by archi- 
tects and engineers; it is the aim of the present book to 
present in as simple a method as is consistent with accuracy, 
the principles which underlie the design of machines and 
structures from the standpoint of their strength. 

The subjects commonly called respectively the Strength 
of Materials and the Theory of Structures have much in 
common; much of the subject matter contained in the 
author's books upon the latter subject has, therefore, been 
Incorporated, the same general method involving the use 
and application of graphical methods in preference to purely 
mathematical methods having been adopted in the other 
branches of the subject. An attempt has been made to 
present more clearly than is general the various theories as 
to the cause of failure in materials and the effect of these 
theories upon design. 

Although the author hopes that the book will be especially 
useful for students reading for the Assoc. M. Inst. C. E., and 
University degree examinations in Engineering, he has 
attempted to present the subject in sufficiently practical 
form for it to be of greater assistance in practical design than 
is the case with an ordinary class book; with this in view 
many diagrams and tables have been incorporated for enabling 
the formulae to be applied with a minimum of time and 
trouble. 

A large number of numerical examples are worked out and 
further exercises are given ; the student is recommended to 



vi PREFACE 

work for himself all such examples and to pay particular 
attention to the assumptions which are made in deriving the 
various formulae. Nearly all engineering formulae are only 
approximately correct ; in the present branch of the subject 
this is chiefly because there is no material known which 
conforms exactly to the simple laws of elasticity upon which 
the subject is based. We cannot condemn too strongly the 
blind application to a particular practical problem of formulae 
which were never intended to be so applied ; the unfor- 
tunate distrust which practical engineers so often have to 
" theory " is to some extent brought about by the fact that 
the theories that they see employed are often inapplicable. 
It is essential for us to acknowledge the limits of theoretical 
methods and not to attempt to express our results to a greater 
degree of accuracy than the nature of the problem will allow. 

The author's thanks are due to Mr. J. H. Wardley, 
A.M.I. C.E., for much assistance and valuable criticism in 
the reading of the proofs; to ]\Ir. W. Mason, D.Sc, for the 
photograph from which Fig. 24 was made ; and to the various 
firms who have courteously assisted by supplying illustrations 
and descriptions of the various testing machines and apparatus 
with which their names are associated in the text. The 
author's indebtedness should also be recorded to the many 
text-books and periodicals that have been consulted and are 
referred to in the various portions of the book. 

The author will be grateful for the notification of clerical 
and other errors that may be found in the book. 

EwART S. Andrews. 

Qoldsmitha* College, 

New Cross, S.E. 

May 1915. 



CONTENTS 

CHAPTER I 

pa(;k 

Stress, Strain, and Elasticity 1 

Kinds of stresses and strains — Hooke's Law — PoiSson's ratio — 
Stress-strain diagrams — Elastic moduli and the relations between 
them — Principal stresses — Ellipse of stress — Maximum strain 
due to combined strains — Resilience — Impact and temperature 
stresses — Stresses in heterogeneous bars. 

CHAPTER II 
The Behaviour of Various Materials under Test . 41 

Properties other than elastic — The cause of failure of material — 
Cast iron — Steel, wrought iron and other ductile metals — Real 
and apparent tensile strength — Liider's Lines — Overstrain — 
Timber — Stone, concrete, cement and like brittle materials — 
EfiFect of relative breadth and height of compression specimens 
— Tensile and shear strength of concrete — Adhesion between 
concrete and steel. 

CHAPTER III 

Repetition of Stresses ; Working Stresses . . 84 

Wohler's experiments — Natural elastic limit — Effect of speed 
upon results — P'atigue of metals — Sudden change of section — 
The conflict between theory and practice — Working stresses 
and factor of safety — Allowance for live loads. 

CHAPTER IV 
Riveted Joints ; Thin Pipes ..... 102 

Forms of rivet heads and joints — Methods of failure — Efficiency 
of joint — Practical considerations — Thin pipes and cylinders — 
The collapse of thin pipes under external pressure. 

CHAPTER V 
Bending Moments and Shearing Forces on Beams . 121 

Cantilevers and simply supported beams ; standard cases — 
Graphical construction — Relation between load, shear and B.M. 
diagrams — A template for B.M. diagrams — B.M. and shear 
diagrams for inclined loads. 

vii 



viii CONTENTS 

CHAPTEK VI 

PAGE 

Geometrical Properties of Sections .... 161 

The determination of areas — Sum curves — Centroid- — Moment 
of inertia and radius of gyration — Momental ellipse or ellipse 
of inertia — Polar moment of inertia — Graphical constructions — 
Formulae for standard cases. 

CHAPTEE VII 

Stresses in Beams 192 

Neutral axis — Assumptions in ordinary beam theory — Moment 
of resistance — Unit section modulus; beam factor — Discrepancies 
between theoretical and actual strengths of beams — Diagonal 
square sections — Influence of shearing force on stresses — 
Moment of resistance in general case. 

CHAPTER VIII 

Stresses in Beams {continued) ..... 215 

Reinforced concrete beams — Straight line, no-tension method — 
General no-tension method — T beams — Combined bending and 
direct stresses — Beams with oblique loading. 

CHAPTEE IX 
Deflections of Beams ....... 248 

General relation in bending — Curvature — Mohr's Theorem and 
graphical investigation of general and standard cases — Mathe- 
matical investigation of standard cases — Resilience of bending. 

CHAPTEE X 
Columns, Stanchions and Struts .... 279 

Buckling factor and slenderness ratio — Methods of fixing ends, 
Euler's, Rankine, Straight line, Johnson, Gordon, Fidler and 
Lilly formulae — Reinforced concrete columns — Braced columns 
— Eccentric loading of columns. 

CHAPTEE XI 
Torsion and Twisting of Shafts 311 

Stresses in a shaft coupling — General case of grouped bolts or 
rivets — Circular shafts ; horse-power transmitted ; solid and 
hollow shafts — Combined bending and torsion — Shafts with 
axial pull or thrust — Torsional resilience — Torsion of non- 
circular shafts — Effect of keyways in shafts. 



CONTENTS ix 

CHAPTEE XII 

PAGE 

Springs 336 

Time of vibration — Close-coiled helical springs — Safe loads on 
springs — Open-coiled helical springs — Leaf or plate springs — 
PivSton rings — Plane spiral springs— Close-coiled helical springs 
under bending stress. 

CHAPTEE XIII 
The Testing of Materials .... . . 364 

Testing machines of various types — Calibration of testing 
machines — Grips and forms of test piece — Extensometers — 
Autographic recorders — Torsion testing machines — Torsion 
meters — Repetition stress machines. 

CHAPTEE XIV 
The Testing of Materials {continued) . . . 396 

Impact, ductility and hardness testing — Arnold ; repeated 
impact and Izod's impact machines — Brinell hardness test — 
Testing cement and concrete — Tension tests ; compression 
tests ; special gravity, fineness, soundness and setting tests for 
Portland cement — Thermal and optical methods of testing 
materials. 

CHAPTEE XV 
Fixed and Continuous Beams 416 

Effect of fixing and continuity — Fixed beams with uniform, 
central and uniformly increasing loads — Fixed beams with 
asymmetrical loading — Continuous beams of two equal spans — 
Effect of lowering central support — Theorem of Three Moments 
— Table of results for equal spans and uniform loads — Graphical 
treatment of continuous beams — Beams fixed at one end and 
supported at the other. 

CHAPTEE XVI 
Distribution of Shearing Stresses in Beams . . 470 

Horizontal shear^ — Distribution of shear in standard cases — 
Graphical treatment — Deflection due to shear-strain of cross 
section of a beam due to shear. 

CHAPTEE XVII 
Flat Plates and Slabs 488 

Slab coefficients — Bach's theory for circular slabs — Grashof's 
results — Grashof-Rankine theory for square and rectangular 
slabs — Bach's theory for square and rectangular slabs with 
various modifications. 



X CONTENTS 

CHAPTEE XVIII 

PAGE 

Thick Pipes 510 

Lame's Theory ; variation of hoop and radial stress ; maxiiauni 
shear stress ; maximum stress equivalent to strain ; diagram 
for designing thick tubes — Shrinkage stresses in a compound 
tube, and strengthening by initial compression. 

CHAPTEE XIX 
Curved Beams . 529 

General analj'sis of stresses — Winkler's formula ; graphical 
treatment ; Resal construction, special cases of rectangular and 
circular section ; correction coefficients — Andrews -Pearson 
formula — Rings and chain links. 

CHAPTEE XX 

EoTATiNG Deums, Disks AND Shafts .... 552 

Thin rotating drum or ring — Revolving disk ; disk of uniform 
strength — Whirling of shafts ; critical speeds ; centrall}'- 
loaded and unloaded shaft ; Dunkerley's approximate 
formulae. 

Exeecises ..... .... 571 

Appendix — Tables of Properties of British 
Standard Sections 

Mathematical Tables 

Index 



THE STRENGTH OF MATERIALS 

Note. — Portions marked with an asterisk may be omitted on the 

first reading. 

CHAPTEE I 

STRAIN, STRESS, AND ELASTICITY 

Strain may be defined as the change in shape or form of 
a body caused by the application of external forces. 

Stress may be defined as the force between the molecules 
of a body brought into play by the strain. 

An elastic body is one in which for a given strain there 
is always induced a definite stress, the stress and strain being 
independent of the duration of the external force causing 
them, and disappearing when such force is removed. A body 
in which the strain does not disappear when the force is 
removed is said to have a permanent set, and such body is 
called a plastic body. 

When an elastic body is in equilibrium, the resultant of 
all the stresses over any given section of the body must 
neutralise all the external forces acting over that section. 
When the external forces are applied, the body becomes in 
a state of strain, and such strain increases until the stresses 
induced by it are sufficient to neutralise the external forces. 

For a substance to be useful as a material of construction, 
it must be elastic within the limits of the strain to which it 
will be subjected. Most solid materials are elastic to some 
extent, and after a certain strain is exceeded they become 
plastic. 

B 



2 



THE STRENGTH OF MATERIALS 



Hooke's Law — enunciated by Hooke in 1676 — states that 
in an elastic body the strain is proportional to the stress. 
Thus, according to this law, if it takes a certain weight to 
stretch a rod a given amount, it will take twice that weight 
to stretch the rod twice that amount ; if a certain weight 
is required to make a beam deflect to a given extent, it 
will take twice that weight to deflect the beam to twice that 
extent. 



<s^ 



E.x7ension 
I 



ac- 



^-iSy-F 



dombr 



u— X. --r 



l?a. 



nsi^erse -i^irain 



^tn 



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Fig. 1. 



RivcT under shear sTrqi/i 
-Kinds of Strain. 



Kinds of Strain and Stress. — ^Strains may be divided 
into three kinds, viz. (1) an extension; (2) a compression; 
(3) a slide. Corresponding to these strains we have (1) tensile 
stress; (2) compressive stress; (3) shear stress. 

A body that is subjected to only one of these, is said to be 
in a state of simple strain, while if it is subjected to more 
than one, it is said to be in a state of complex strain. 

Examples of simple strains are to be found in the cases of 
a tie bar; a column with a central load; a rivet. The best 



STRAIN, STRESS, AND ELASTICITY 3 

example of a body under complex strain is that of a beam 

in which, as we shall show later, there exist all the kinds of 

strain. 

Intensity of Stress. — Imagine a small area a situated 

at a point X in the cross section of a body under strain, then 

if S is the resultant of all the molecular forces across the 

S 
small area, - is called the intensity of stress at the point X. 

In the case of bodies under complex strain, the intensity of 
stress will be different at different points of the cross section, 
while in a body subjected to a simple strain, the stress will 
be the same over each point of the cross section, so that in 
this case if A is the area of the whole cross section and P 
is the whole force acting over the cross section, the intensity 

p 

of stress will be equal to . . In future, unless it is stated 

to the contrary, we shall use the word " stress " to mean the 
" intensity of stress." 

Unital Strain. — The unital strain is the strain per unit 
length of the material. In the case of extension and com- 
23ression, the total strain is proportional to the original length 
of the body. Thus, a rod 2 ft. long will stretch twice as 
much as a rod 1 ft. long for the same load. In Fig. I, if 
I is the unstrained length of the rods under tension and 
compression and x the extension or compression, the unital 

strain is j . 

In the case of slide strain, the angle but not the length of 
the body is altered, and this angle /3 is a measure of the 
unital strain. If the angle is small, as it alwaj^s will be in 
practice with materials of construction, then it will be nearly 

equal to j, where x and I are the quantities shown on the 

figure. 

Poisson's Ratio — Transverse Strain. — When a body 
is extended or compressed, there is a transverse strain tend- 
ing to prevent change of volume of the body. The amount 



4 THE STRENGTH OF MATERIALS 

of transverse strain bears a certain ratio to the longitudinal 

strain. 

rr^i . ^. transverse strain . ^ 

Ihis ratio = , .— ,^ — p- — . = -n varies from i to i for 

longitudinal strain ' ^ * 

most materials, and is called Poisson's ratio. 

According to one school of elasticians, the value of this 
ratio -q should be J, but experimental evidence does not quite 
support this view, although it is ver}^ nearly true for some 
materials. The ratio is very difficult to measure directly, its 
value being best obtained by working backwards from the 
elastic moduli in shear and tension in the manner which will 
be explained later. 

Stress-strain Diagrams. — If a material be tested in 
tension or compression, and the strain at each stress be 
measured, and such strains be plotted on a diagram against 
the stresses, a diagram caUed the stress-strain diagram is 
obtained. If a material obeys Hooke's Law, such diagram 
will ^be a straight line. For most metals, the stress-strain 
diagram will be a straight line until a certain point is 
reached, called the elastic limit, after which the strain in 
creases more quickly than the stress, until a point called the 
yield "point is reached, when there is a sudden comparatively 
large increase in strain. After the yield point is reached, 
the metal becomes in a plastic state, and the strains go on 
increasing rapidly until fracture occurs. 

Fig. 2 shows the stress-strain diagram for a tension speci- 
men of mild steel, such as is suitable for structural work. 

The portion a b of the diagram is a straight line, and 
represents the period over which the material obeys Hooke's 
Law. At the point c, the yield point is reached, and the 
strain then increases to such an extent that the first portion 
of the diagram is re-drawn to a considerably smaller scale, 
such portion being showni as ab^ c^. The strain then increases 
in the form shown mitil the point d is reached, the curve 
between c^ and d being approximately a parabola. When 
the point d is reached, the maximum stress has been reached. 



STRAIN, STRESS, AND ELASTICITY 5 

and the specimen begins to pull out and thin down at one 
section, and if the stress is sustained, fracture will then 
occur. The portion d e, shown dotted, represents increase 
of strain with apparent diminution of stress. This diminu- 
tion is only apparent because the area of the specimen 



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Fig. 2. — Stress-strain Diagram for Mild Steel. 



beyond the point rapidly gets smaller, so that the load may 
be decreased, and still keep the stress the same. In practice, 
it is very difficult to diminish the load so as to keep pace 
with the decrease in area, so that this last portion of the 
curve is very seldom accurate, and has, moreover, little 
practical importance in commercial testing because the 



6 THE STRENGTH OF MATERIALS 

maximum stress is always taken as that given at d (see 
also p. 52). 

The specimen draws down at the point of fracture in the 
manner shown in the diagram. Before the test, it is cus- 
tomary to make centre-punch marks at equal distances apart 
along the length of the specimen. The distance apart of 
these points after fracture of the specimen indicates the 
distribution of the elongation at different points along the 
length. Four such marks, a, h, c, d, are shown on the figure. 
The greatest extension occurs at the point of fracture, so that 
on a specimen short length, the percentage total extension 
will be greater than on a longer specimen. We deal further 
with this point on p. 55. 

The stress-strain diagrams in compression and shear for 
mild steel are very similar to that for tension. In compres- 
sion it is difficult to get the whole diagram, because failure 
occurs by buckling, except on very short lengths, where it is 
very difficult to measure the strains, and in shear the test 
is best made by torsion, because it is almost impossible to 
eliminate the bending effect. Now, in torsion, the shear 
stress is not uniform, so that the metal at the exterior of the 
round bar reaches its yield point before the material in the 
centre, and this has the effect of raising the apparent yield 
point. We shall see later that the same occurs in testing for 
compression or tension by means of beams. 

The importance of the elastic limit has been overlooked to 
a great extent by designers of machines and structures ; but 
inasmuch as the theor}^ on which most of the formulae for 
obtaining the strength of beams are based, assumes that the 
stress is proportional to the strain, it must be remembered 
that our calculations are true only so long as Hooke's Law is 
true, so that the elastic limit of the material is a very impor- 
tant quantity. We shall deal further with this question in 
discussing working stresses (Chap. III.). 

Confusion between Elastic Limit and Yield Point. — In 
commercial testing, it is quite common to use no accurate 



STRAIN, STRESS, AND ELASTICITY 7 

means for measuring the strains (instruments for such 
measurements are called extensometers , see pp. 371-379). The 
load on the steelyard of the machine is run out until the 
steelyard suddenly drops down on to its stops. The " steel- 
yard-drop " happens when the yield point is reached, but 
many people call this the elastic limit ; it is also sometimes 
called the " apparent elastic limit." As will be seen from the 
diagram, Fig. 2, there is no appreciable error made by this 
confusion in tension testing, but in cross bending the differ- 
ence is much more marked, and gives rise to confused ideas. 
We shall deal further with this point on pp. 207-209. 

The Elastic Constants or Moduli. — If a material is 
truly elastic, i. e. if the strain is proportional to the stress, 
then it follows that the intensity of stress is always a 
certain number of times the unital strain, or that the 

, . intensity of stress . . . tvt . i • . , • 

ratio 7—^1 — : — -. is constant. Now this stress-strain 

unitai strain 

ratio is called a modulus. That for tension and compres- 
sion is generally known as Young's modulus, and is given 
the letter E ; that for shear is called the shear, or rigidity 
modulus (G). There is an additional modulus called the hulk 
or volume modulus (K), which represents the ratio between 
the intensity of pressure or tension and the unital change in 
volume on a cube of material subjected to pressure or tension 
on all faces. 

Young's modulus is the one with which we shall be most 
concerned. Suppose that a tension member (a tie as it is 
called) or a compression member (a strut), of length I and 
cross-sectional area A is subjected to a pull or thrust P, and 
that the extension or compression is x, Fig. 1. Then the 

intensity of stress is -j, and the unital strain is y- 

P X PZ 

.* . Young's modulus = E = -r ^ -r = * — 

^ A I Ax 

Numerical Example. — A mild-steel tie bar, 12 ins. loyig 
and of \\ ins. diameter, is subjected to a -pull of IS tons. If the 
extension is '0094 in., find Young's modulus. 



8 THE STRENGTH OF MATERIALS 

Area of section of 1| ins. diam. = 1*767 sq. ins. 

18 
.-. Stress per sq. in. = ..^pr- = 10-19 tons per 

' ' sq. in. 

0004 
Unital strain = .^ = 000783 

.-. Young's modulus = ^-^^ = 13,000 tons per 
^ -000783 sq. m. 

The value of Young's modulus can be found from the 

stress-strain diagram. Thus, in that for mild steel, Fig. 2, 

E = -i 

X 

-r stress 

Xow in the relation E = - ! , if the strain is equal to 1, 

strain ^ 

i. e. if the bar is pulled to twice its original length, we have 
that E = stress, and this accounts for the definition of 
Young's modulus that some writers have given, viz. : Young's 
modulus is the stress that is necessary to pull a bar to twice 
its original length. Some students find this definition more 
clear than the one previously given, but it must be remem- 
bered that no material of construction will pull out to twice 
its original length without fracture. 

Relation between Elastic Constants. — For an elastic 
material there will be certain relations between the elastic 

moduli E, G, K, and Poisson's ratio — . These relations can 

be found as follows — 

To first find a relation between E and K consider a cube 
of unit side subjected to a pull ff, Fig. 3 (a). 

Let the elongation along the axis be a, and let the trans- 
verse contraction be b. 
Then original volume of cube = 1 

strained volume of cube = (1 + a) (1 — 6)- 

= 1-26 + 62 a-a-2o/;-a62 

= I -T- a — 2b (nearly) 
because as the strains are ver}' small their product ma}' be 
neglected. 

.• . Increase in volume = (l + a — 2 6) — 1 

= (a - 2 6) 



STRAIN, STRESS, AND ELASTICITY 



9 



Now apply a pull to each side of the cube. There will now 
be three pulls, each joroducing an increase of volume equal 
nearly to {a — 2 6). 




/I 



D 



>f) 





Fig. 3. 

Total increase in volume is nearly equal to 3 {a — 2 h) 

2 b' 



3a 1 - 



a 



AT ^ _ transverse strain _ 
a longitudinal strain 



10 THE STRENGTH OF MATERIALS 

. • . Increase in volume = 3 a ( 1 — 2r}) 

.' . Since original volume = 1 

increase in volume , . ^ . ■ « /i « ^ 

^-. — J — -, = volume unital strain = Sail — 2'n) 

ongmal volume 

,, intensity of pull /^ 

Now K = -T— I'S^ — • — = '5 — n — o~T 

unital strain S a [l — z-q) 

, ft tensile intensity of stress ^^ . , , 

and ' = ^— 1^ r,-^— 7 — -. = Young s modulus 

a unital tensile strain 

= E 

E 

•*• ^ = 3(1 -2-r,) ^^^ 

Now find the relation between E and G as follows — 

Suppose that two shearing forces of intensity s are applied 

to the faces of a unit cube a b c d, Fig. 3 (6). Now consider 

the equilibrium of the portion a d c, Fig. 3 (c). To balance 

the forces s there must be a force pulling /, along the diagonal 

A c, and the value of f, must be V2 x s. Now the area over 

which this force acts will be V2 since the cube is of unit 

Y^2 s 
side, so that there will be a tensile stress of - .— = s. 

V2 

Similarly considering the portion BCD, Fig. 3 {d) there 
must be a compressing force /, along the diagonal b d, and 

■\/2 s 
the compressive stress will be = , ^ s. Therefore we 

see that : Two shear stresses on planes at right angles to each 
other are equivalent to tensile and compressive stresses of in- 
tensity equal to that of the shear stress at right angles to each 
other, and at an angle of 45° to the shear stresses. 

Now the cube will be deformed to the shape AiBjCiDi, 
Fig. 3(e). 

The unital shear strain is measured by the angle of distor- 
tion 2 </). Since the strains are very small, this is practically 

, , ^ B Bo Z B Bo . . 1 \ A 

equal to -^ ^^ = — ^ (since b c = 1) = 4 b B2. 

2^ B C 2 

Let the unital strain due to the tension along the diagonal 
B D be a. Then there will also be a strain along this 



STRAIN, STRESS, AND ELASTICITY 11 

diagonal due to the transverse strain from the compression 
stress in AC. This will be equal to r; x a. .' . Total unital 
tensile strain along diagonal = a (1 + ry). Then b Bj = unital 
strain along diagonal x J b d, since b b^ is equal to d d^ 

V'2 a 

.• . B Bi = a (1 + ^) X O B = ^— (1 + -q). 

Because the strains are in reality very small, b Bg b^ is 
very nearly a right-angled triangle. 

.'. b Bi = ^2 X B Bg 

B Bt a {\ -\- 7}) 

or B Bo = —T^- = — ^ ^ " 
^ V2 2 

-p. intensity of tensile stress _ ft _ ^ 
unital tensile strain a 



, intensity of shear stress _ s 
unital shear strain 4 b Bg 



= G 



Since we have proved that the tensile stress along the 
diagonal is equal in intensity to the shear stress. 

Therefore /, = aE=Gx4BB2 

E _ 4BB.2 _ 4a (1 + 7;) 
' ' G a 2a 

^ = 2(l + >?) (2) 



• • G 

Now we have already shown in (1) that 

E 



K 


3(l-2.y) 


From (2) rj = 


^ - 1 

2G 


From (3) y = 


1 E 

2 6K 


A _ 1 _ 
2G 


1 E 


2 6K 


2VG^ 3k; 


3 

' 2 


1 1 


3 


G"^ 3K ~ 


E 


9 


3 1 


or ^^ = 


^. + , 


E 


G ^ Iv 



(3) 



(4) 



12 THE STRENGTH OF MATERIALS 

This expresses the relation between the constants in its 

simplest form. 

It will be noted that if ry = |, as some authorities state, 

E 
then ^ = 2-5; this may be taken as true if the value of G 

for the material is not known. 

Strains in different Directions.^ — Suppose that the 
stresses in a material acting normally to, i. e. at right angles 
to, three planes at right angles to each other are /^, f,„ /.. 
Then the unital strain in the first direction is made up of 
the direct strain due to /j. and the transverse strains due 
to the other two. 

.*. unital strain in first direction = 5, = 1^ — ' ^^ " ^ 

E E 

/ yj (f -L. f ) 

unital strain in second direction = s„ = J. — -^ ''' 

E E 

unital strain in third direction = 5, = (i — ^ ^1' 'y' 

E E 

Lateral Strain prevented in one Direction. — Now take 
the case of a piece of material which is free to dilate or expand 
in one direction, but is prevented from doing so in another 
at right angles, a compression stress /^ being applied in the 
third direction. 

Let the z direction be that in which strain is prevented. 

We then have 

and let the .t direction be the one with the stress /^. 

Then if /. is the stress caused in the z axis we have, since 
i,j = 0, because the material can expand freely in the y 
direction — 

E s, ^ j, - q j. 
* E ^, = - ^ (/. 4- /O 

0=/.-r;/,. 



.-.E^, = -^(1 + .^)/,; -Es, = JA^-T) 



s. [l-v) 



STRAIN, STRESS, AND ELASTICITY 



13 



This result is interesting as showing that the ordinary 
definition of E only holds when lateral strain is not 
prevented. 

* Complex Stress. — Principal Stresses. — It can be 
shown that Avhen a body is under a complex system of 
stresses, such stresses will be the same as those due to the 
combination of three simple tensile or compressive stresses in 
planes at right angles to each other. Such simple stresses 
are called the principal stresses. 

Consider the case of a block of material subjected to pulls 
P and Q, Fig. 4, in two directions at right angles, and let 



Q 



^ 




-B 




Fig. 4. — Principal Stresses. 



the pull ]jer square inch of the sectional area in each direction 
be p and q, respectively, these being principal stresses. 

Consider the stresses on a plane a b inclined at an angle 
to the force P. 

The stress p can be resolved perpendicularly and along 
A B, i.e. normally and tangentially to a b. 

Now consider 1 sq. in. of area perpendicular to p. The 



corresponding area along a b will be 



1 

sin 0' 



Now the component of p perpendicular to A b wiU be p 



14 THE STRENGTH OF MATERIALS 

sin 0, and the component along a b will be p cos 0, but 
stress = component of force -^ area. 

.• . Normal or perpendicular component of stress p along a b 

= p sin - — : — 7 = p sin^ 
^ sni ^ 

tangential or shear component of stress }j along a b 

= V cos -^ . ,, = p sin cos 
sm ^ 

Now considering stress q, its tangential component to a b 
will be opposite in direction to that of p, and since in this 

case the area is -. — j^^^^ — = - — - and the normal and 

sm (90 — 0) cos 

tangential comj)onents of q are respectively q cos and q 

sin 0, the normal comjoonent of stress will be q cos- 0, and the 

tangential component of stress will be — g sin cos 0, since 

in this case the tangential components are not in the same 

direction. 

.* . Total normal component = /„ = p sin"^ -{- q cos^ . . (1) 

Total tangential component ^ s = {p — q) sin cos . . (2) 

Now the resultant of the stresses /„ and s, which we will call 

/, will be given by a c. 

i.e./ = V"/;^ + 52 

^ V {p sin^~0^~Y^os^^~f'+'{p-qf sin^ 6* cos^ 
= V p^ (sin^ + sin2 cos^ 0) + q^ (cos^^m^T+cos^) 
+ 2pq (cos^ sin^ — cos^ sin^ 0) 

— V p^ sin^ ^ (cos^ + sin^ ^) + q^ cos^ ^ (sin^ + 

cos2 ^) + 

- V /?2 sin2 + q^ cos'^ ^ (3) 

because cos'^ + sin^ ^ = 1. 

The inclination a of this stress is given b}' 

/„ p sin^ + Q cos^ d 

tan a = ^^ =- ; -X-- — zl - ^ 

5 ip — q) sm 6^ cos f^ 

= ?L:^_^1^1^ + ^ (4, 

[p—q) tan ^ 



STRAIN, STRESS, AND ELASTICITY 15 

If <f) is the angle between a c and the direction of p, 

Then cf, ^ {a - 0) 

tan a - tan 6 
tan> = tan {a - 0) = i ^ ^an a . tan ^ 

PJ^^^JAJ _ tan e 
_ {p — q) tan ^ 

~ . , p tan^ 6 + q . ^ 
1 + f r^r — I . tan ^ 

_ p tan^ ^ + g — (P — g) t^iJ-^ ^ 
~ {p — q) tan ^ + tan {p tan^ ^ + g) 
_ g (1 + tan^J) _ q 
~ p tan ^ (I + tan^ $)~ p tan ^ 

= ^cote (5) 

p 

Unlike Stresses.— If the stresses are unlike, i.e., one 

tension and one compression, we shall have by similar 

reasoning 

f„ = p sin^ ~ q sin^ 

s = [p -{- q) sin cos 

It will be noted that for p =' q and ^ = 45° we have /„ == 
and s == p. 

We have, therefore, a pure shear as equivalent to equal 
tensile and compressive stresses at right angles to each 
other, at 45° to the shear stress and equal in intensity to 
the shear stress (cf. p. 10). 

The Ellipse of Stress. — Draw circles of radius o x and 
o Y, Fig. 5, equal to q and p respectively, and let o R be 
drawn at angle ^ to o y. 

Draw a radius o f to the larger circle at right angles to 
o R and cutting the smaller circle in e. 

Draw F H at right angles to o y, and e g at right angles to 
F H, and join o G. 

Now o H = o F cos (90 — 0) = p sin 6 
and G H = E K = o E sin (90 — ^) = q cos 
. • . o G = V o H^ + H G^ = V p^ sin^ + q^ cos^ 6 
.• . by equation (3) o g = / 



16 THE STRENGTH OF MATERIALS 

Now tan h O G = i^r = • . /, = cot ^ = tan <f) 



HG a cos U Q . , 

= ^ . „ = ^ cot i 

OH 'p sin 6 p 

.• . / H o G = ^ and, since a =^ ~\- (fi, / g o b, = a. 

Now the locus of the point G is an ellipse of major axis 
2 p and minor axis 2 q, and such ellipse is called the Ellipse r ' 
Stress. 

We see, therefore, that by drawing a line o f from the 

y 5LLIP3E OF STRESS. 



R 





F/ 


^ G 




H 
K 


/ 




A 




/ 


:\ Y^- 


/ 






v£^ 


^^/ 




Y 








\ 


-A. 


k 




V^ 


be 




K 




> 



/•5- 



*^ 



M 



Fig. 5. — Ellipse of Stress. 

centre o, at right angles to a given direction to the outer 
circle, and drawing f h horizontal to meet the ellipse of 
stress in g, then o g gives the resultant stress on a jDlane 
in the given direction, and the angle g o r = a gives the 
angle between such resultant stress and the plane. 

Numerical Example. — Suppose a square bar of 2 ins. side 
and 4 ins. long is subjected to pulls of 10 and 12 tons respectively 
in axial and transverse directions. Find the resultant stress on 



STRAIN, STRESS, AND ELASTICITY 



17 



a plane inclined at 60 degrees to the axis, and find the inclination 
of the stress to that plane. 

In this case V = ^ =" 2'5 tons per square inch, 



V 



and q 



4 
12 



1*5 tons per 'square inch. 



Then Fig. 5 shows the elhpse of stress drawn to scale. 

Draw o L at 60° to o Y and draw o m at right angles to o l 
to cut the outer circle in m ; drawing m n horizontal to meet 
the ellipse of stress m n, then o n gives the resultant stress 



fS. 




Fig. 6. — Combined Normal and Shear Stress. 

and / L o N gives its inclination to the plane, o n will be 
found to be 2' 29 tons per square inch, and / l o n to be 79°. 

Now considering again the stresses p and q and the normal 
and tangential stresses f„ and s at an inclination 9 to p we 
see that p and q are the principal stresses corresponding to 
the stresses /„ and s. Now in practice we often require to find 
the magnitude and inclination of the principal stresses, because 
one of these stresses will be that of maximum intensity of 
stress. This is clear from the figure of the ellipse of stress, 
since o Y is obviously the maximum radius vector of the 
ellipse. We will now therefore find the principal stresses due 
to a normal stress /„ and a shear or tangential stress s at right 
angles to each other. 



18 THE STRENGTH OF MATERIALS 

* Combined Normal and Shear Stress. — To investi- 
gate this problem Ave must first j^rove that a shear stress must 
always be accompanied by an equal shear stress at right 
angles to it. Take for example a unit cube, Fig. 6 (a), sub- 
jected to shearing forces S along opposite sides. These forces 
S form a couple, and the cube can be kept in equilibrium 
only by another couple of equal moment and opposite sense, 
which couple is given by shearing forces S^ at right angles to S. 

Now consider the case of a complex system of stress con- 
sisting of a normal stress / and a shear or tangential stress s. 

Let p N, Fig. 6 (6), represent a portion of the plane on 
which the stresses / and s act. 

Let one of the planes of principal stress be represented by 
p M, and let this principal stress be p. Then along m n there 
acts a shear stress also of intensity s. 

Then the resolved portions of the forces due to p and to the 
stresses / and s must be equal in the directions p n and m n. 

Therefore we have 

/ . P N + 6- . M N = ;/J . P M COS (1) 

also 5 . p N = /> . p M sin (2) 

„ , , , P N M N 

. • . Jb rom ( 1 ) / ' + s =^ p cos 

^ ' ' PM p M ^ 

i. e. / cos 6 -}- s sin 6 = p cos 

• ' • iV ~ f) cos 6 = s sin (3) 

P N 

From (2) s = p sin 6 

^ ' p M ^ 

.' . s cos = p sin (4) 

.*. Dividing (3) and (4) we have 

P - / _ « 
s p 

p {p — f) = s^ 

P" — p f — s^ =^ 



-^=i(i±Vi+f) (5) 



STRAIN, STRESS, AND •ELASTICITY 11) 

The minus sign corresponds to the second principal stress q, 
which will be in compression ; as we are concerned only with 
the maximum stress, we will take the positive value, viz. — • 

^-K^+V^") (^) 

The direction of the plane at which this stress occurs is 

given by 6. This is found as follows — 

From (3) j) cos — f cos = s sin 6 

From (4) p sin ^ s cos 6 

s cos 6 
.' . p =^ 

sin 

S COS" u ! p. • /I in\ 

.' . . ,, — / cos = s sni 6 ( / ) 

sm 6^ ' 

.' . s (cos^ — sin- 0) =^ f sin cos 

, sin 2 ^ 
.' . s cos 2 6 = f — ^— 

or tan 2 ^ - ^- (8) 

This will give two values of 0, 90° apart, and so gives the 
inclination of both planes of principal stress. 

Maximum Shear Stress. — Returning to the consideration 
of the principal stresses p and q, we saw that the tangential 
component on a plane at angle to p was given by {p — q) 
sin cos 6. (See p. 14, equation (2)). Now this will be a 

maximum when sin cos ^ is a maximum, i.e. when — ^ — 

is a maximum, or when = 45°. Therefore we see that the 
maximum shear stress occurs at 45° to the principal stresses, 

and is equal to ^^^— «— • 

In the problem that we are considering, we have proved that 

^ = K^ + a/i + ^) ^""^ ^^^^ ^ = 2 (^ " a/i + ^^' 



2 ~ 2 V ^ /2 



20 THE STReJv^GTH OF MATERIALS 

. • . Maximum shear stress = ^ . i ^^^ (9) 



OJ- = \/ J+^' "(10) 



\4 



The latter form is more convenient because in the case when 
/ = 0, the former gives an indeterminate result. 

Numerical Example. — A steel bolt, 1 in. in diameter, is 
subjected to a direct pull of 3000 lbs. and to a shearing force of 
1 ton. Find the maximum tensile and shearing stresses in lbs. 
per square inch, and the inclinations of the directions of the 
stresses to the longitudinal axis of the bolt. 

T ,1 . , 3000 3000 

in this case / = . - . ~^.t = --^^ . 

area of 1 m. bolt '7854 

= 3819 lb. per sq. in. 

s = .>y^w = 2852 lb. per sq. in. 

. • . Maximum tensile stress = p =^ ' i\ -\- ^J \ -\- --- 

3819 /, . /,' . 4 X 28522\ 



{^-^^-^^) 



= 5342 lb. per square inch. 

Inclination of jmncipal plane to plane perpendicular to 

axis is given by 

, ^ . 2 5 2 X 2852 
^^^^^^=J- 3819- 

= 1-494 

.-. 2^ = 56° 12' nearly 
.-.^ = 28° 6' 
.• . Inclination to longitudinal axis = 90 — 28° 6' 

= 61° 54' 



STRAIN, STRESS, AND ELASTICITY 

? 



Maximum shear stress = 



+ s' 



38192 



A/ "^ 



+ 28522 



- 2852 y 1 + 4 >,-2852^ 

= 2852 V 1 + -448 

= 2852 X 1-203 

= 3428 lb. per square inch. 



21 



■ 


'i, 


N 


s 




i 




A 




JoC' 






d 





Fig. 7. 



This stress will occur at 45° to the direction of principal 
stress, i. e. at 61° 54' — 45° == 16° 54' with longitudinal axis, 

or else at 90° to this, i. e. at 73° 6' with longitudinal axis. 



* Combined Shear Stress and two Normal Stresses 
at Rig-ht Angles to each other. 

Next consider the case of a shear stress s combined with 
normal stresses /^, f,, at right angles to each other (Fig. 7). 
Then resolving as before we have 

/:,?N + 5MN =2^.PM cos (11) 

/,/ N M + <5 P N = ;p . P M sin ^ (12) 



22 THE STRENGTH OF MATERIALS 

From (11) /., cos -\- s sin --- p cos 
From (12) /, sin ~\- s cos = p sin 

i. e. {p — f^) cos ^ = 5 sin ^ (13) 

{p — f,) sinO =- s cos 6 (14) 

We therefore have by multiplying and cancelling sin 6 cos 6 

[p - U) iP - h.) - ^' 
i.e. p^^-p (/. + /,) + /.,. /, = ^2 • (15) 

Solving this quadratic we get 

P = H(/' + /.") ± V1/.. + /,)— 4177/7^^)} 



ih + fy)^ /(/.-/.) 



2 ±V''^T^+'' ^'^^ 

* In this case as in the previous one we usually take the 
positive sign. 

To get the direction of the principal stresses we have from 
(13) and (14) 

{p - /^.) = s tan^ (IV) 

ip-fy) =scote (18) 

.*. subtracting (18) from (17) 

(/. -fy)-^ (cot - tan 0). 

__ ^ _, 2tan^ 2 _ 2 

Now tan 2 ^ = i-tan ^ ^ ZZT^a^O ~ ^"^ " *^^ ^ 

tan 

i.e. ^ot0-t^n6 = ^^^^ 
• • ^'^ '^'' tan -2 6* 

29 

z.e. tan2^= , _ , (19) 

\h I III 

[p — q) 
The maximum shear stress is as before equal to ^ — 

• . Maximum shear stress = ^J '" . — h 5^ 

Graphical Representation of Results.— The following 



STRAIN, STRESS, AND ELASTICITY 



23 



graphical construction, due we believe to Professor R. H. 
Smith, solves these equations. 




Fig. 8. — Graphical Construction for Combined Stresses. 

Set out o A, Fig. 8, to represent /;, to a convenient scale 
and o B to represent fy to a convenient scale ; if /^ and fy are 
opposite in sign they should be set out in opposite directions. 



24 THE STRENGTH OF MATERIALS 

Bisect A B in c and at B set up b d at right angles to o a 
to represent the shear stress s ; then with centre c and c d 
as radius draw a semicircle, cutting o a in f and e. 

Then o f = ^ and o e = g. 

If E comes on the other side of o, the stress is negative. 

Join D F, then /d f b = 6, the angle of the principal stresses. 
Also maximum shear stress = c e. 

Proof,— b c = ^2^ = f^^' 



C D = Vb C2 + B d2 = j(t_^ + 5^ 
.'. OF = OC + CF = OC+CD 



(/. + /.) , Kh-h) 



2 

OE = OC — CE = OC — CD 



\2 



+ ^r-'^^r^ + s^ = p 



CE = C D = 



2- - V 4 "^^ -^ 

if — f )^ " 

o ~ + 5^ = maximum shear stress. 



Now /b c D = angle at centre = 2 d f b 

, B D s 25 

2 
. • . From equation (19) b c d = 2 ^. . * . /d f b = ^. 
Application to a Single Normal Stress. — In this case, 

Fig. 9, B and o coincide so that oc = |oa = ^, and the 

construction comes as shown. 

This figure has been drawn for / = 3819 and 5 = 2852 as 
in the numerical example of p. 20. 

* Maximum Strain compared with Maximum Stress. 
— In questions involving complex stresses it is necessary to 
remember that the maximum strain does not occur on the 
same plane as the maximum stress. There is some con- 



STRAIN, STRESS, AND ELASTICITY 



25 



siderable divergence among elasticians (a term suggested by 
Professor Karl Pearson, F.R.S.) as to whether the ultimate 
criterion of strength of a material depends on the tensile 
or compressive stress exceeding a certain value, or the shear 
stress exceeding a certain value, or on the strain exceeding a 
certain value. This is dealt with on pp. 42-48. 

We have considered the cases of maximum tensile or 
compressive and shear stresses. We will now consider the 
question of maximum stress. 




Fig. 9. 



The given stresses are equivalent to simple stresses j), q 
at right angles to each other; we will assume p to be 
greater than q* 



. . Strain in direction of p 



P_vq 

E E 



[t] = Poisson's ratio). 



Simple stress in direction of p to cause same strain 
= Equivalent direct stress p, = p — -q q 

rjf 



= 2(i + V^ + 7? 



2 \ 



1-A 1 + 



45' 

7= 



2 {{1 - V) +:(1 + v)^Jl + *f } (11) 



26 



THE STRENGTH OF MATERIALS 



In the direction at right angles there will be an equivalent 
stress equal to q — y p which comes equal to 



/ 



/ )(l_,,)_(l-f-^)yia-4f| 



(12) 



These formulae may also be derived from first princijiles 
as follows. 

Suppose a rectangular block a b c d receives two tensile 
strains at right angles and a slide strain in the same plane. 

Under the combined strain the block assumes the position 



^,--^^- 



Ci 



i ' c 




' ^ 


^ ■ 


1 y^^ 






y 


' y^ 




:Xe 




'y^ 





ac 



1^, 



I / I 



I 

I ' 
I / 
I / 
I / 
I' 



B 



■ " "B, "iz 



Fig. 10. — Combined Strains. 

A Do Co B^. Then, if ab = .r, b c = y, and a c = r, and 
Xi yi Ti are the strained lengths, and /Dj a d., = /? 

-i' J' 

Unital strain in direction .r = s^ = — 

X 

yx - y 



y 



_ 'I 



. • . We have x\ ^ x (1 - -5,) (1) 

(2) 

(3) 



yi = //(!- s.) 

r^ = r (1 -h s,) 



= rH\ 



-^s,: 



Since squares of strains may be neglected. 



(4) 



Now r^ = A Cg^ = A Bg^ + C2 B 2 



STRAIN, STRESS, AND ELASTICITY 27 

A B2 + C2 B2 
= (A Bi + Bi B2)2 + A Di2 
= (ABi + DiD2)2 + ADi^ 

Now A Bj = a^i = X (1 + Sx) 
A Di = 2/1 = y (1 + 5,) 

since ^ is small and therefore ^ x 5^ is of second order and 
therefore negligible. 

.-. ri2= {x(l+5.) +^2/}' + {2/(1 +^.)!' 

= x'{\ + 2s^) + 2xy/3 + y^l +2s,),...{5) 

neglecting all second powers of strains, 

but r^ = x^ + y^ 

.-. rj^^ = r^ + 2x^8, + 2y''-s, + 2xy /3 (6) 

. • . From (4) 

r2 (1 + 2 s,) = r^ + 2 x^ s, + 2 y^ s, + 2 X y f3 

or..= g)%.+ (f)%,+ ^2^^ (7) 

Expressing this in terms of the angle we get 

S0 = Sr cos^ 6 + s,, sin^ 0^/3 sin cos ^ (8) 

Our next problem is to find the value of 6, for which the 
resultant unital strain Sq is a maximum. 

This occurs when ,-/ = 
a $ 

i.e. when 

s^ . 2 cos d ( — sin e) + Sy 2 sin a cos a + j8 (cos cos + sin [ — sin 6]) = 

i. e. when — Sr sin 2 ^ + 5^^ sin 2$ + /3 cos 2^=0 

sin 2e{s, - Sy) = ^ cos 2 6* 



w 



or tan 2$ = — ^ — ..(9) 



This gives two values of at right angles, and so we see that 
the directions of maximum strain are at right angles. 

Now consider equation (8), reuniting and putting 1 = cos^ 
+ sin^ 0, we get 

S0 (cos^ + sin^ 0) = s,. cos^ + Sy sin^ 6 + (3 sin $ cos $. 



28 THE STRENGTH OF IMATERIALS 

Dividing by cos^ 6, we get 

Se (1 + tan^ 0) = s^ -\- s,, tan^ -{- /S tan 
or tan- 6 {s,j — Se) + /3 tan 6 + s^ ~ Se = 

I.e. tan Q - — ^{s,-s^ — 

For this to be real, 

/32 must be not <C 4 (5y — 5^) {s^ — se) 
Now as S0 increases, 4 {Sy — se) (5^. — Se) will increase, since 
the latter expression is equal to 4 {so — s^) {se — s,j) 
. ' . The greatest value 5^ can have is such as to make 
)82 = 4 {s, — S9) {s^ — Se) 

i. e. se'^ — Se {s^ + s,) + s^ Sy — ^ = 
, e. ., = ^+_^_^V_|: -JE+F (10) 

Now consider the first case for which we have worked out 
the principal stress, viz. the combined stress due to a tensile 
or compressive stress / and a shear stress s. (Note. — This 
shear stress s must not be confused with the strains s„ etc.) 
In this case if 5^ = strain due to stress /, the only strain in 
direction y is the transverse strain due to s^, i.e. Sy = — -q s^ 
(negative because the transverse strain is compressive). 

Considering only the positive value in equation (10) 

. . se = —2 

/ s 

Now 5^ = p and (S = ^ 

Also Se = &, where p, is the equivalent principal 

stress due to considering the maximum strain, E and G 
being the Young's and shear moduli. 

■■E~ 2E '^ 2"S E^^'^^' G^ 
but|-2(l + ,;) 



STRAIN, STRESS, AND ELASTICITY 29 

Now rj is very nearly ^ for steel. 

. • . taking this value, we get 



^'-2(i+iV^+y) (12) 

Comparing this with the corresponding equation (6) 
(p. 19), from considering the stress we see clearly the 
difference between the results from the two points of view. 

Numerical Example. — Consider the same problem as 
worked on p. 20. 

In that case / = 3819 lbs. per square inch. 

s = 2852 „ 



Pe 



3819/3 ^-^Jl + i. 


X 28522\ 


2 U 4 V + 


38192/ 


^f(| + |V3-23) 




^2^^ (-75 + 2-246) 


* 


^^^^^ X 2-996 





= 5722 lbs. per square inch. 

To get the inclination at which the maximum strain occurs 
return to equation (9) by which 

tan 2 ^ = ^ 

S.r S,, 



In this case we get 



tan 20 = 



s s 

G G s . E 



sAl + i) f{^+v) /(1 + V7).G 
E 



_ 5.2(1 + 7]) _ 2s 
f{l + vr~ f 
This is the same as in the case considering the principal 
stress, and so has the value as given before. 



30 



THE STRENGTH OF MATERIALS 



Similarly for normal stresses /^ and /„ at right angles we 
have 

■"^ ~ E ~ E , 
^^ E E 



/? = 



G 



s. + s, = ^^ ^ ^^ it + /,) 



s,- — s, = 



E 

(1 +V) 
E 



(/. - /.) 




Fig. 11. 



.*. From equation (10) 

, _ ^^ _ ih + h) (1 - v) . /(^-J..)i(i_+_i)' 4. i' 

' E "2E - V 4E ' "^4G2 

= ^^[{L + h) {l-v)± ^J{f. - f.r (1 + v') + ~^-} 
= 2 E !(/^ + /^) (1 - ^) ± (1 + ^) V{f.-f,)' + 4.s^\ 

Ve = \ {(/. + /,) (1 - >/) ± (1 + ^) V(/.^ W^TIT^J . . (13) 

* Shear Strain equivalent to Two Direct Strains at 
Right Angles. — We will now consider from first principles 



STRAIN, STRESS, AND ELASTICITY 31 

ill a similar manner the shear strain equivalent to two direct 
strains at right angles to each other. 

Let a rectangular block a b c d, Fig. 11, become strained 
to the form a b^ c^ d^, the extensions being shown to a 
greatly exaggerated scale, then we have 

AB, = y {I + s,) (14) 

A Bi = X (1 + S,r) (15) 

.*. Neglecting squares of strains 

A Cj^ = A Bi^ -f A Bj^ 

= X2 (1 + S,,)^ + 2/' (1 + 5,)2 

= x^ -h y^ -h 2x^ s,,. -\-2ifs,, (16) 

= (a'2 + 2/2)(l + 2 5,) 

=. a;2 + 2/2 + 2 a;2 s. + ^y^-s, (17) 

.♦ . A Ci2 - A E22 = 2 X'2 {S,, - 5,) (18) 

Now A C^^ — A Eg^ = (a Ci + A Eg) (A C^ — A Eg) 

= 2 A Eg . Ci Eg approx. 

^ X [S,. S,i) t -t r\\ 

.•. Ci Eo = ^ '-' 19) 

^ ** 2AE2 



Further Eg Eg = c^ Eg tan 6 (very nearly ; strictly 

X y {s,, — s„) 



tan^- 8$) = CjEg^ 

X 



. * , Eg Eg 



AE 



but SO = 2 — ( = — ^. — approx. ) 
A Eg \ radms ^ ^ / 



AEg^ 

{s, - Sy)xy 
(l + 25,)(x2 + 2/') 






32 THE STRENGTH OF MATERIALS 

This will be a maximum when + " is a minimum, i. e 

y X 

tan $ + cot ^ is a minimum. 

sin , cos 6 sin^ d + cos^ S 



tan + cot 



cos 6^ ' sin sin 6^ cos 6 
1 2 



J sin 2 ^ sin 2 
The minimum value of this = 2 when sin 2^ = 1, i.e., 
(9 = 45°. 

.• . Maximum shear strain occurs at 45° to direct strains. 

CJ ^x S,i • 

■ • ' ~ Y{rv2s;) 

= ' ^ ' (1-2 5,) approx. 

= ^—^ — ^neglecting products of strains. 

There will be an equal angular distortion on the other 
diagonal. 

.' . Total angular distortion = shear strain = 28^ = {s^ — s,j). 
. • . Equivalent maximum shear stress 

= shear strain x ,G = (<s^ — s,,) G. 
Now let the principal stresses be p and q ' ' 



en 


s, = 


' E 


~ E 




s,= 


E 


E 


.-. {s. 


-^J- 


iv- 


- 9) (1 + -n) 
E 


• •• {Sx- 


5jG = 


Av- 


-g)(l+r7)G 



- ^^ ^^ because E - 2 G (1 -f q) (p. 11). 

.*. Equivalent maximum shear stress = -- 

It will be noted that the equivalent and actual maximum 
shear stresses come the same, whereas the equivalent and 
actual principal stresses are different. 



STRAIN, STRESS, AND ELASTICITY 



33 



Resilience. — The work done per unit volume of a material 
in producing strain is called resilience. Consider the case of a 
body subjected to a simple tensile strain. In going from the 
point A to the point b. Fig. 12, very near to it, the average 
stress acting is /. Therefore, if a b = x, the work done by 
the force / in straining the material from the point a to the 
point B will be equal to / x x. Now, if x is the increase in 
unital strain and / is the intensity of stress, the volume of 
material acted upon is unity. Now, a B is assumed to be 
very small, and f x x is equal to the area of the shaded 
portion of the stress-strain curve. 




Fig 12. — Resilience. 

Therefore, the resilience is equal to the area of the stress- 
strain curve up to the point m, 

i.e. resilience = area of A p m x 

_ 1. 

— 2' ^ "^ 



Now, — = Young's modulus = E 



resilience in tension ^ 



2E 



similarly in shear the resilience = ^^ 

where s is the shear stress. 
Stresses and Strains due to Sudden or Dynamic 
Loading. — If a load is applied suddenly to a structure, 



D 



34 



THE STRENGTH OF MATERIALS 



vibration will ensue, and the strain — and thus the stress — 
will reach twice the value which would occur if the load 
were gradually applied. 

This will be made clear from considering a diagram, Fig. 13 
(1), where the force is plotted against the strain. We have 
seen that, with gradual loading of an elastic body, the curve 
representing the relation between the strain and the load in 
direct stress is represented by a straight line a d, the area 
below the line giving the work done up to a given point. 




Fig. 13. — Sudden or Dynamic Loading. 

Now let A G represent a force P ; then when the strain gets to 
the point b, the work done by the force will be equal to the 
area of the rectangle a b e g, whereas the work done in 
straining the material is only equal to the area of the triangle 
a B E, so that there is an amount of work equal to the area of 
the triangle a e g still available for causing increased strain. 
The strain therefore increases until the area of the triangle 
E F D is equal to that of the triangle A e g. It is, clear that 
AC = 2 A B, or that the strain — and thus the stress — is twice 
that in the case of gradual loading. 

If a force is suddenly reversed from — P to + P, then the 



STRAIN, STRESS, AND ELASTICITY 



35 



total strain and stress will be the same as that due to a 
sudden load of 2 P, and again when the strain reaches the 
point B, Fig. 13 (2), there will be an amount of work repre- 
sented by the area of the triangle a E g still available for 
causing strain, which therefore continues to the point c. 
Thus the maximum tensile strain will be equal to H l. If 
the loading were gradual the strain would be h k, and as 
H L = 3 H K, we see that a load suddenly reversed causes three 
times the strain and stress which occur if such reversal takes 
place slowly. 




jc , 




'ratn . 



Fig. 14. 



Fig. 15. 



In each of these cases the additional strain or stress which 
occurs is equal to the amount of variation. Such additional 
stress has been called the dyrw^mic increment, and we there- 
fore see that the equivalent gradual stress due to a sudden or 
dynamic stress f,, which varies by an amount v is given by 

h + v. 

strain and Stress due to Impact. — Suppose a weight W 
falls from a height h on to a structure and let the deformation 
or strain in the direction of h be x, Fig. 14. Then the work 
done by the weight is equal to W {h + x). Now this Work is 
absorbed in straining the structure. Consider first the case 
in which the resulting strain is within the elastic limit. The 
work done in such case is equal to the volume multiplied by 



36 THE STRENGTH OF ]\L4TERIALS 

the resilience. We have shown that in tension or compression 
the resilience is equal to — -^ and therefore in this case we get 

W (/i +0;) = ^ -pT — ^ "^ 9i?- Then if x is negligible 

compared with h 

we have W x 7i = ^ -ni 



If the weight strikes with a velocity v, 

^9 



2E.Wv^ /EW 

or 






2?V -"V^V 

We will consider resilience in bending and torsion when 
dealing with beams and shafts. 

Strain beyond Elastic Limit. — If the strain is beyond 
the elastic limit, it follows, from the reasoning given on 
p. 33, that the work done per unit volume in straining is 
equal to the area below the stress-strain curve. If this area 

is R, Fie;. 15, then we have R ^ W /^ or ^ — 

° 2 g 

From this the stress can be found. 

Numerical Example. — A bar of \-inch diameter stretches 
\ inch under a steady load of 1 ton. What stress would he 
'produced in the bar by a weight of 150 lbs. which falls through 
3 inches before commencing to stretch the rod — the rod being 
initially unstressed and the value of E taken as 30 x 10^ lbs. per 
square inch. {B.Sc. Lond.) 

Area of bar J" diam. = -196 so[. in. 

.-. Stress under load of one ton = .iq/:> tons per sq. in. 

2240 „ 
= Tjge lb. per sq. m. 

. _ Stress _ 2240 

•' • ^^^^"^ ~ "E ~ -196 X 30 X 10« 



STRAIN, STRESS, AND ELASTICITY 37 

Now ^" = strain x original length 

r. - • ^^ 4.1. ^ '196 X 30 X 10« 

.*. Original lenefth = ?^~-.- = ^^.^ r. 

^ ^ Strain 2240 x 8 

.' . Volume = length x area of section. 

_ -196 x 196 X 30 X 10 « 

8 X 2240 

. = 64'33 cub. ins. 

Work done by 150 lbs. in falling 3 inches 

= 3 X 150 = 450 in. lbs. 

64-33 X /2 



= 450 

/9( 

^| 64-33 



2E 

r _ /900lE 



900 X 30 X 106 



64-33 

= 20,480 lbs. per sq. in. Ans.* 
Temperature Stresses. — Suppose a bar of length I is 
heated t° F. and a is the coefficient of expansion. Then, un- 
less prevented, the length of the bar will become I (1 x at), 
i. e. the increase in length will he at I. 

If the bar is rigidly fixed so that this expansion cannot take 
place, then there will be in the bar a strain equal to at I, and 

the unital strain will be -j— = at. 

This strain will produce a compressive stress of a ^ x E, 
where E is Young's modulus. 

Now for mild steel a = -00000657 per degree Fahrenheit, 
and E = 13,000 tons per square inch. 

.-. The stress per °F. = '00000657 x 13,000 

= -0854 tons per square inch. 

Taking a range of temperature of 120° F., the stress due to 
temperature = 120 x -0854 = 10*25 tons per square inch. 
This is more than the safe stress for mild steel, so that the 
importance of designing so that the expansion may take 
place becomes quite evident. 

* This problem could be solved if E were not given; it would be 
found to cancel out. 



38 



THE STRENGTH OF MATERIALS 



"^ Heterogeneous Bars under Direct Stress. — If a 
bar, composed of two different materials — such as steel and 
concrete, or steel and coj)per — firmly connected to each other, 
be subjected to a pull or a thrust, the two materials must be 
strained by equal amounts, and since the values of Young's 
modulus for the two materials are different the stresses in the 
two materials will be different. 

Suppose one material has a cross-sectional area A and 
Young's modulus E, the resulting stress being / ; and let the 
corresponding quantities for the other material be A^, E^, Z^. 

Then, if iinder a pull or thrust P the unital strain is a*, we 
have 

I 
E 

h 

El 
and P = A / 



X = 



X = 



(1) 



Ai/r 



(2) 
(3) 




Fig 16. 



A / and Aj ii being the loads carried by each of the materials. 

E,/ 



Prom (1) and (2) /^ = Ej^x 



.-. P = /(A- 



E 

El A, ) 
E ' 



or / = 



Afl^^tV 



(4) 



(5) 



Now if a new bar is taken wholly of the first material of 
such area A., that the stress under a load P is the same as 
that in the compound bar, we have 

P 

A2 

El A, ^ 



/ 



or A2 = A ( 1 -T- 



EA 



(6) 



/A = — ^ (8) 

^ + EA 



STRAIN, STRESS, AND ELASTICITY 39 

This quantity Ag may be called the equivalent area of homo- 
geneous material, and the consideration of this problem has 
become in recent years much more important on account 
of the progress made in reinforced concrete construction. 

Returning to the general problem we see that 

p 

The load carried hj the first material then comes equal to 

P 

EA 

and that carried by the second comes equal to 

/iAi = --Va (9) 

Since these are not the same there will be an adhesive 
force tending to make one material slide relatively to the 
other. 

This adhesive stress may be computed as follows assuming 
that the load is applied uniformly. 

Load per sq. in. = — r — ■ — r— 

(A + Ai) 

. • . Load actually distributed to area A 

- A ^ 

(A + A,) 

p 

Load carried = t^ . (from 8) 



EA 

E 

Difference = load carried by adhesion calling ^ = 

_ P P. A 

'"'-, ,_A, A + Ai 

^ + mA 
_ p J m A A 

"" ( m A + A'l ~ Ai + A 
_ P A A i (m - 1) 

~ (A + Ai)XArrf wA) 
_ / iAAi( m -^1) 
(A + Aj) 



m 



40 



THE STRENGTH OF IMATERTALS 



Numerical Example. — A reinforced concrete column 
{Fig. 17) for which m = 15, is 20 inches square and has 4 
1^" steel rods embedded in it. Find the load on the column 
when the stress in the concrete is 450 lbs. per sq. in. and the 
adhesive force. 

T 




Fig. 17, 



TT 



4 X ^ X 1-252 = 4-91 in. 
4 



In this case A 

Ai = 20 X 20 - 4-91 = 395 nearly 
.-. P 






= 450 X 395 1 + 



15 X 4-91 



Adhesive force 



395 

210,870 lb. nearly 
45 X 4-9 1 X 395 xU 
400 



(from 7] 



= 33,940 lbs. nearly 



CHAPTER II 

THE BEHAVIOUR OF VARIOUS MATERIALS UNDER 

TEST 

Properties other than Elastic. — In addition to the 
elastic properties of materials there are other strength 
properties which are of very great importance in the 
practical use of the materials. 

Ductility is the property of a material which allows it to 
be worked without cracking ; the strict use of the term refers 
to the capacity for being drawn out which a ductile metal 
possesses. 

Malleability is the property which allows a material to be 
hammered out and is very similar to ductility. 

Brittleness is lack of ductility or malleability. 

Hardness may be defined as the power of a material to 
resist denting by another material. (For tests for hardness 
see pp. 396-404.) 

The above properties are all relative ones and vary with the 
same material according to the treatment which it receives ; 
thus by "tempering " a metal we harden it and by " anneal- 
ing " it we soften it or render it more ductile, and some 
metals are hardened by plunging them into water when 
heated, whereas others are annealed by the same process. 

With reference to ductility it is important to remember 

that so long as a metal maintains its elasticity it has no 

ductility, i.e.au metal which possesses ductility cannot exhibit 

the fact until the yield point has been reached. In our 

calculations for the strength of various details we shall base 

41 



42 THE STRENGTH OF MATERIALS 

nearly all our formulae on the assumption that our material 
is elastic and so we must not expect the formulae to hold 
after the elastic Hmit has been reached. This is a point of 
very great importance. 

Apart from the convenience in the manufacture of articles 
which ductility gives, it has considerable value from the point 
of view of safety and strength, because a material does 
not lose its strength when it first starts drawing out, and 
the yield may either give us timely warning of excessive load- 
ing or, in the case of steam boilers and like fluid-retaining 
devices, the yielding may actually remove the excessive 
pressure. As we shall see later, however, the effect of taking 
a material beyond its yield point is to harden it. 

The usual test for ductility is the elongation in fracture 
by tension. 

* The Cause of Failure of Materials under Stress. 
— In recent years a very large amount of attention has been 
given to the question of the cause of failure of materials 
under test, and it is doubtful if the vital importance of this 
problem has been fully realised by practical engineers. As we 
shall see later, however, the choice of safe working stresses 
really depends in a large measure upon the view taken as to 
which of the various theories is correct. If we consider the 
question carefully we shall see that failure cannot occur by 
compression only ; if a material be prevented from escaping 
laterally, no amount of compression can rupture it. Even a 
fluid like water will resist a compression stress of very great 
magnitude if the vessel containing it is strong enough to resist 
failure by tension or shear. The late M. Armand Considere 
showed experimentally that concrete could not be crushed 
when given adequate lateral support, and he also proved that 
the very brittle material glass could be bent cold without 
fracture when placed in a liquid under great hydraulic pres- 
sure. Marble has been bent without fracture by Professor 
E. D. Adams of McGill University when placed in steel cylin- 
ders and compressed, and Professor Ira H. Woolson crushed 



BEHAVIOUR OF MATERIALS UNDER TEST 43 

a cylinder of concrete encased in steel into the form shown 
in Fig. 18, and yet when the encasing cylinder of steel was 
removed the strength of the concrete was found to be not 
appreciably different from that of concrete which had not 
been similarly treated. The question therefore resolves itself 
whether tension or shear is the cause of failure, and we have 
reason to believe that in ductile materials such as mild steel 
failure occurs by shear and in brittle materials such as 
cement or concrete by tension. We will return to this after 
considering the various theories of failure ; there are four 
principal theories which we will consider. 

1. Principal Stress or Rankine Theory. — ^According to 




Fig. 18. 

this theory, which was adopted by the great Glasgow professor, 
Rankine, the failure occurs when the maximum principal stress 
exceeds a certain value. We have seen (p. 19) that for a 
normal or direct stress / and shear stress s the principal stress 
is given by the relation 



/ , 1 , 

or ?9 = -^ + 2 V/2 + 4^2 (16) 

and the inclination of this stress to the normal stress and 

to the shear stress is given by the relation tan 20 = —r. 

This stress p is the simple normal stress (tension or com- 
pression) equivalent in effect to the combined normal and 
shear stresses. 



44 THE STRENGTH OF MATERIALS 

In the limiting case in Avhich the direct stress / is zero we 
get p = s and tan 2 $ = infinite, i. e. = 45°, i. e. a shear 
stress is equivalent to a normal stress of the same intensity 
and is at 45° to it, or the shear and tensile strengths of the 
material should be equal. 

2. Principal Strain or St. Venant Theory. — According 
to this theory, which was favoured by the great French elas- 
tician after whom it is named, the failure occurs when the 
maximum principal strain exceeds a certain value. We have 
seen (p. 29) that for a normal stress / and a shear stress s 
the equivalent principal stress is given by the relation 



P>- 



L\a -yi) + a +-n)J} 4- ^^^\ 



2 



j^(l->y) + (l +'/?)yi + y, j- (2) 



and taking rj = I 



/ f 3 , 5 / 4^1 
^• = 2i4 + 4V^+fJ ^^^^ 

orp = ^J + g V/^ + 4^2 (36) 

The inclination of this principal stress is the same as in the 

previous case. 

In the limiting case in which the direct stress / is zero we 

5s . 
get p ^ -, i.e. a stress shear is equivalent to a normal stress 

of four-fifths of the shear stress, or the shear strength of a material 
is four -fifths of the tensile strength. 

3. Equivalent Shear Stress or Guest or Tresca 
Theory. — According to this theory, which is associated with 
the name of Mr. J. J. Guest, who was one of the first to 
carry out careful experiments upon the subject, and is usually 
attributed to Tresca, failure occurs by sliding of the particles 
over each other, i. e. by shear. From p. 20 we get 

Equivalent shear stress = . ' +52 ^4^ 

and acts at an angle of 45° to the normal stress. 



BEHAVIOUR OF MATERIALS UNDER TEST 45 

To compare a simple shear with a simple tension or 
compression by this formula we put s = o and we get 

Equivalent shear stress = ~ 

i. e. a shear stress is equivalent to a normal stress of twice its 
magnitude or the shear strength of a material is one-half of its 
tensile strength. 




Fig. 19. — Navier's Theory. 

It is interestmg to note that as shown on p. 32 the 
equivalent shear stress comes the same whether worked from 
the point of view of stress or of strain, and so there is logical 
support for this theory. 

4. Sliding with Internal Friction or Navier Theory. 
— This theory deals with materials subjected to compressive 
stresses and attempts to explain the fact that short cylinders 



46 THE STRENGTH OF MATERIALS 

of brittle material usually fail by sliding or shearing along 
a line e f, Fig. 19, inclined at an angle from 55° to 65°, on 
the ground that the particles are capable of exerting frictional 
resistances. 

Consider a short column a b c D of unit sectional area 
subjected to an ultimate compressive force u,, which causes 
failure, and consider the forces across a section E f, the 
ultimate or breaking shear stress in the material being u,. 

The force u,. acting along the section e f can be resolved 
into shear and normal components ac, cb respectively, equal 
to u, sin and u, cos 0. 

If /x is the angle of friction for the material, the normal 
component c b causes a frictional resistance equal to /x .c b, 
i.e. /x .u cos 6. 

Just before failure the shearing force acting upon E r = 
u^ X area of section. 

_ u, X normal section _ u, 
cos cos 

(because normal section is of unit area). 
When therefore failure is about to take place — 
Total force causing failure = ac = tc^ sin equals 

Total force resisting failure = au, cos -1 ^ 

* ^ cos 

i.e. u, (sin — ix cos 0) = ' 

^ ' cos Q 

or n, = „ , . -.' r. ••••(!) 

cos Q (sm B — \j. cos b) 

Regarding w, and \}. as constant we now wish to find the 
value of which will make u, as small as possible ; this value 
of will be that at which failure will occur. ii will be as 
small as possible when cos B (sui — ix cos 0) is as large as 
possible. 

Let y — cos (sin — /x cos 0) 

sin 2 ^ , . 

= —IX cos- 



BEHAVIOUR OF MATERIALS UNDER TEST 47 



T-i . d y ^ 

I<or a maximum -7^ = 

a 6 

2 cos 2 ^ ^ . , ... ^ 

^. e. ^ — fjL 2 cos 6^ (— sm 6') = 

i. e. cos 2 -j- jx sin 2 6^ = 

cos 2 ^ 

/A = ^. — ^--7, = — cot 2 ^ . . (2) 

sm 2 ^ ' 

If ft = tan <^, ^ being the angle of friction, this gives 

cot 20 = — tan </> 

or 2 ^ - 90° + <^ 



or == 45° + 



(3) 



In support of this theory experiments made by Bouton 
(Washington University, 1891) may be quoted as follows — 



Material. 


Number 
of Tests. 


Observed 
Value of (f) 
(degrees). 


Observed 
Value of e 
(degrees). 


4r + ^ 

55-3 
53-4 
61-7 

58-6 
58-5 


Cast iron 

„ (different kind) 
Limestone 
Asphalte paving 
Milwaukee brick 


24 
24 
4 
3 
4 


20-6 
16-9 
33-4 
27-3 
27-0 


54-8 
55-0 

62-2 
59-7 
58-2 



Batio of shear to compressive strength on Navier theory. — 
From equation (1) we have 

Us = u, cos (sin — fx cos 0) 

Now put in this a = tan (h = — cot 2.^ = v~^ ^ 

^ sm 2 

rp, . . Us . / . . , COS 2 (9 .^ 

I his gives = cos ^ sm ^ + -^ — ?r^cos 
^ Ur \ sm 2 (9 



= cos 



^ ( . ^ , (cos^ — sin^ 0) cos 0} 

6' ] sm ^ + -^ 5-^ — n -^ } 

[ 2 sm cos J 

. r . , , (cos2 - sin2 0)] 

\ sm + ^ o-^—n } 

[ 2 sm J 

f 2 sin2 + cos2 - sm^ 0] 

' ^ 1 2^nl / 



— cos 

cos ^ / . \ COS 

2 sm ^ \ / 2 sm 

- I cot ^ (4) 

Taking 6^ = 60 for masonry, this would give 

shear strength = '289 compressive strength, 



48 THE STRENGTH OF ]\L\TERIALS 

but the most recent experiments on shear strength of concrete 
(see p. 79) indicate that this is too low. Shear tests are 
very difficult to make without introducing bending, which 
tends to give the shear strength too low. 

An interesting theory which may be regarded as a 
modification of Navier's theory is outlined by i\Ir. H. 
Kempton Dj^son in a paper read before the Concrete Institute, 
December 1914. 

The first three theories have been very fully tested experi- 
mentally in recent years by Messrs. Guest, Hancock, Scoble, 
C. A. H. Smith, Mason and Turner,* and the result appears 
to be that the shear stress theory is most reliable for ductile 
materials while the strain theory is most reliable for brittle 
materials. 

Professor Ewing and ^Ir. Rosenhain have found by a 
microscopic examination of the crystals of a ductile metal 
under strain that beyond the yield point lines of slip are 
developed in the crystals, thus proving that the failure or 
yield is a slipping or shear one. Liider's lines (p. 54) are 
also indications that in ductile metals the failure is by shear. 

It is possible that ductility is a property of shear strength ; 
if a material is weaker in shear than in tension the shear 
causes the failure and slippage occurs before the tensile strength 
is reached, thus giving rise to ductility. If the material 
is relatively stronger in shear than in tension the material 
breaks before slippage occurs and thus cannot exhibit ductility. 

We will now consider the properties of various materials. 

CAST IRON 

We will deal with cast iron first because it is a brittle material 
and behaves differently under test from most other metals. 

The strength of cast iron varies considerably with its com- 
position, but like all brittle metals it is relatively weak in 
tension and strong in compression. 

* These experiments will be found fully described and discussed in 
various articles and letters in Engineering for 1909 and 1910. 



BEHAVIOUR OF MATERIALS UNDER TEST 49 

Fig. 20 shows the stress-strain diagrams for cast iron in 
tension and compression, the resiiHs of the two tests being 
plotted on one diagram. 

It is clear from this diagram that the stress is never strictly 
proportioned to the strain in tension ; this has an important 
bearing upon the strength of cast-iron beams (see p. 209). 

The compression diagram is not continued to failure as the 




Fig. 20. — Stress-strain Curves for Cast Iron. 

failure would take place by buckling and injure the instru- 
ment for measuring the strain. When a cast-iron bar fails 
in tension, it breaks off " short," i. e. it does not produce a 
waist as indicated in Fig. 2 for mild steel. 

When compression tests are made on cylinders which are so 
short that buckUng effects are practically eliminated, the 
failure takes place by sliding diagonally as indicated in 
Fig. 21, and for shorter specimens still cracks sometimes 
develop which split off the outside portion leaving two inverted 
cones. Some observations on this kind of failure for concrete 

E 



50 



THE STRENGTH OF MATERIALS 



will be found on p. 69 and apply to cast iron. The diffi- 
culty in testing cast iron in compression in very short 
lengths is that it is so strong that difficulties arise as to the 
strength of the testing machine. 

The tensile strength of cast iron varies from about 7 to 15 
tons per sq. in. in extreme cases, but more usually from 8 to 11 
tons per sq. in. Figures for the compressire strength show 
more variation ; this is probably due to the fact that the size 
of the test piece, both as regards its length and breadth, affects 
the result. 




Fig. 21. — Compression Failure of Cast Iron, 

The following results of tests made upon J in. cubes by 
the American Foundrymen's Association show that specimens 
cut from bars of small cross section give much higher results 
than those from large. 



Cross Section 


Crushing 


strength in 


tons per sq. 


m. for cubes cut from 


of bar 
from which 






















Cubes 


Middle 


First 


Second 


Third 


Fourth 


were cut. 


half inch. 


half inch. 


half inch. 


half inch. 


half inch. 


4x J 


69-0 


___ 










1 X 1 


44-5 


49-8 


— 


— 


— 


Hx li 


37-0 


39-4 


37-0 


— 


— 


2x2 


32*2 


38-9 


34-6 


— 


— 


2ix 2i 


31-9 


35-4 


32-3 


31-9 


— ■ 


3x3 


28-6 


32-5 


30-1 


28-7 


— 


3i X 3* 


28-4 


30-5 


29-6 


28-8 


28-4 


4x4 


25-4 


29-4 


27-4 


26-6 


25-4 



BEHAVIOUR OF MATERIALS UNDER TEST 51 

Strength of Cast-Iron Beams. — Cast-iron when tested 

in bending shows an apparently greater strength than when 

tested in pure tension. This is due to the fact, as explained 

in greater length on p. 209, that the ordinary formula for 

beams is not strictly applicable for cast iron. The ratio of 

calculated breaking stress from bending . „ , ^ i r- p 

^ — ., , ^ , . is usually about 1*5 tor 

tensile breakmg stress 

rectangular sections of depth twice the breadth ; it increases 

for round and square sections arranged diagonally and may 

rise to 3 ; the ratio becomes nearly 1 for I sections with a 

thin web. 

The following figures give the mean of a large number of 

tests made by Kirkaldy for the same kind of iron — 

Tensile breaking stress = 11 tons per sq. in. 
Compression ,, =54 ,, „ 

Calculated bending ,, =17 „ „ 

The early writers often called the breaking stress calculated 
from bending tests the " modulus of rupture," but the term 
is not to be recommended ; bending breaking stress is better. 

Effect of Temperature on Strength of Cast Iron. — 
The strength of cast iron increases slightly as the temperature 
is raised until about 900° F. is reached; it then diminishes 
rapidly, until at 1100° F. the strength is reduced by nearly 
50 per cent, and at 1400° E. by about 75 per cent. 

Other properties of cast iron are tabulated on p. 83. 

Malleable Cast Iron. — Cast iron is rendered malleable 
by surrounding the casting with haematite or manganese 
dioxide and exposing it to red heat for many hours, depend- 
ing upon the size of the casting. The result of the process 
is to dicarbonise the iron and render it similar to mild steel. 
Some useful information on the subject, especially from the 
point of view of strength, is given by Mr. C. H. Day in the 
American Machinist of April 21, 1906. 

The following results of tests of Mr. Ashcroft are quoted 
from Vol. CXVII. Proc. I. C. E. 



52 



THE STRENGTH OF MATERIALS 







Tons per sq. in. 


Elongation, 
^ on 10 in. 


Breaking Stress. 


Elastic Limit. 


Young's Modulus. 
11,620 


Tension . . 
Compression 


20-6 
21-6 


8-94 


2-8 





10,240 


— 


Bending . . 


28-8 


12,330 ! — 


Torsion . . 


26-8 


— 


Rigidity Modulus. 
4,120 


— 



Stanford, in Trans. Am. Soc. C. E., 1895, gives as the mean 
result of forty -two tests in tension an elongation of 6" 61 per 
cent, and an ultimate stress of 22 tons per sq. in. 

Similar results are given by Mr. Day. 



STEEL, WROUGHT IRON, AND OTHER DUCTILE 

METALS 

Real and Apparent Maximum Tensile Strength. — 

We have shown already, on p. 5, a stress-strain diagram 
for mild steel in tension and pointed out that the last portion 
D E of the diagram was usually inaccurate and of little com- 
mercial importance. The diagram sloi)es back because the 
load can be reduced as the area diminishes and the stress 
still be sufficient to cause fracture. Now this diagram can 
be corrected if its form is determined very carefully and the 
areas at the various points are measured. 

Taking any point a, Fig. 22, on the curve before the 
specimen began to draw down, we find 6 c by the relation 
ab X extended length 
original length 
tion that the volume keeps constant; so that 

original area x original length 



be 



This is based on the assump- 



reduced area x extended length 



ab X extended length 
original length 



ab X actual reduced area 
original area 



BEHAVIOUR OF MATERIALS UNDER TEST 53 

The method cannot be used beyond the point at which the 
waist begins to form. 

Taking any point / beyond the point of drawing down, 



(U 

(!) 




sh 



rc^\n 



b G 

Fig. 22. 



d 



^. csc^ r> 1 7 dfx actual area of bar , , , . 

h 12. 22, we find a e = —^ ^-. — ^ : and by doing 

° origmai area "^ ° 

this for a number of points we get the corrected curve 

c e e', then f e' gives the real maximum stress as opposed 

to G D, which is the -apparent or commercial maximum 

stress. It is very difficult to get points / accurately. 



54 THE STRENGTH OF MATERIALS 

It thus appears that the actual maximum stress at failure 
is considerably more than usually measured. Upon the Guest 
theory that failure occurs by shear, the tensile strength 
should be twice the shear strength, but ordinary commercial 
tests indicate that it is about 1 J times ; it is probable that 
if the true tensile strength were compared instead of the 
apparent or commercial strength, the agreement would be 
more in favour of the Guest theory.* 

The common form of fracture of a ductile metal is shown 
in Fig. 23 and consists of a kind of crater, the angle of the 
sides being approximately 45°. This strengthens the theory 
that failure is by sliding or shear. 

Liider's Lines. — When a highly polished specimen or one 




Fig. 23. — Tension Failure of Ductile Metal. 

provided with a very thin layer of scale is tested in tension 
or compression, lines at about 45° to the axis of the speci- 
men and of spiral form in the case of round sections are 
found to develop directly the yield point is reached. This 
was first noticed by Liider and supports the theory that the 
yield is really due to diagonal shearing. 

Fig. 24 shows these lines for thin mild-steel tubes in com- 
pression and were given in a paper by Mr. W. Mason, M.Sc, 
of Liverpool University. f 

Percentage Increase in Length and Decrease in 
Area. — The percentage elongation of the specimen and the 
decrease in area are usually regarded as reUable tests of 
ductility of the material, but it is clearly useless to specify 
the percentage elongation unless the diameter and the 

♦ See a paper by Professor Carus Wilson, Proc. Roy. Soc. 1890. 
t Proc. Inst. M.E. 1909. 




Fig. 24. — Luder's Lines. 



[To face page 54. 



BEHAVIOUR OF MATERIALS UNDER TEST 55 

original length are both specified, because most of the ex- 
tension occurs in the centre portion where the specimen 
draws down. 

This question has been treated very fully by Professor 
Unwin in Vol. CLV., Proc. Inst. C. E., and the following 
figures are taken from his paper as indicative of the general 
results — 

RoTHERHAM Steel Boiler Plate (area -5830 in.^) 
Extension in inches in each half inch. 



Number 

from 
Fracture 


7 


6 


5 


4 


3 


2 


1 


Frac- 
ture 


1 2 
•19 |'13 


3 
•12 


4 
•12 


5 
•11 


6 


7 


8 


9 
•10 


10 
•09 


Extension 


•07 


•08 


•10 


•10 


•11 


•13 


•18 


•47 


•11 •lO -09 



Percentage elongation and different gauge 


lengths. 




Gauge Length (inches) 
% Elongation . . . 


2 


4 
36^0 


6 


8 


10 


48^5 


30^9 


27-6 


25^9 



He suggests the formula — 



cVA 



% elongation = 100 {pi^ + h\ 



where A is the original area in sq. in. 
I is the gauge length in inches 
h and c are constants for a given material. 

The following values of c and h are given by Professor 
Unwin — 



Metal. 


c 


6 

18 
10-6 
9-7 
•8 
35 


Mild steel 

Gun metal (cast) . . . 
Rolled brass .... 
Rolled copper .... 
Annealed copper . . . 


70 

8^3 
101-6 

84 
125 



56 THE STRENGTH OF IMATERIALS 

By the formula we can obtain the jorobable percentage 
extension on any length of any other metal if that on two 
specified lengths are known. 

The percentage contraction in area is not so commonly 
specified now as formerl}^ because the elongation is considered 
as giving sufficient indication of the ductility. 

Two Waists ix a Texsiox Specimen. — It happens very 
occasionalh^ that two waists form in a tension specimen, 
failure taking place at the one which draws down most 
rapidh'. This will affect the elongation and the test should 
be discarded. 

Fracture xear one Shoulder of Specimen. — If fracture 
occurs near one shoulder of the specimen (see Fig. 167) the 
elongation will be less than normal owing to the effect of 
the shoulder, and such a test should be discarded. 

Effect of Abrupt Change of Section upon the 
Tensile Strength. — The effect of an abrupt change of 
section such as in a screw thread or a sharp-edge groove 
does not have a ver}' marked effect upon the ultimate 
tensile stress of a material Avhen tested in the ordinar}^ way 
although it does affect the ductility. As we shall show later, 
however (p. 91), it has considerable effect in tests by 
repeated loading. 

The sharpness of the groove will have a weakening effect, 
but the presence of the larger area near it will have a 
strengthening effect . 

Fig. 25 shows the results of some tests by Sir Benjamin 
Baker * which are interesting ; the breaking stresses in 
tons per sq. in. are given below each figure. a is an 
ordinary tension specimen, h and c have saw cuts, at both 
and one side respectively, d has semicircular notches and 
e has a central hole with saw cuts at each side; after the 
saw cuts were made the bars were heated and the cuts 
closed, d is the strongest ; this would be expected, on 
the shear theory, as the diagonal line is relatively larger; 
* Proc. Inst. C. E., Vol. LXXXIV. 



BEHAVIOUR OF MATERIALS UNDER TEST 57 



b is weaker than d on account of the abrupt change of 
section ; c is weaker still because the load is not central, and 
in e the hole probably gives an accentuation of the effect of 
the abrupt edges. Eor other results on the effects of holes 
see Chap. XIV. 

Effect of Overstrain. — If a ductile metal is loaded 
beyond the j'ield point and the load removed, and the speci- 
men is then loaded up again at once, it is found that the 
new yield point is higher, but the elastic limit is slightly 
lower. The overstrain also increases the ultimate or break- 
ing tensile stress. This is shown in Fig. 26 (a) in which b 



cc 



L 



d 



^nV~7V~7^^ 



32.-50 



Lji\!_i 



-o 



3l-^0 



Fig. 25. 



36 '30 



L_J] 



is the initial or primitive elastic limit, and a is the initial 
yield point; at the point c the load is taken off and then 
the specimen is loaded up almost immediate^; the new 
elastic limit e is much lower than previously and the yield 
point d higher. If some hours had been allowed to elapse 
between taking off the load and reloading, the elastic limit 
would nearly return to its previous value b but the yield 
point would go higher still to the point g. 

In Fig. 26 (6) is shown the effect of keeping the load on 
for some time at the point c before increasing it further. 
The curve in full lines shows the effect of keeping the load 
jfixed for about ten minutes, and in dotted lines k I the effect 
of keeping it for ten days.* 

* For fuller information a paper by Professor Ewing, Proc. Roy. 
Soc. 1880, should be consulted. 



58 



THE STRENGTH OF MATERIALS 



This hardening effect of overstrain is well known in prac- 
tical work. Copper wire becomes very brittle by bending it 
backwards and forwards, and steel wire in the process of 
drawing becomes very hard indeed. 













K 












?5 








( 
1 


^/ 






k 


f'^' 


I 






^ 


^""^ 


^ 






,^ 




c 


'/' 


20 


c^y 


^ 








^l/ 


/ 










b 










h 










15 






























e 














10 












































5 



































































o 



5 



10 



o 



5 

(h) 



/o 



Fig. 26. — Overstrain. 



Recovery of Elastic Limit from Overstrain. — As in- 
dicated above, the elastic limit slowly recovers its original 
value after it has been allowed to rest for a few hours ; it 
then will increase as the time of rest is extended and 



BEHAVIOUR OF MATERIALS UNDER TEST 59 

ultimately gets above its final value and gets near to its 
new yield point. 

Mr. Muir * has shown that the temperature of boiling water 
gives an almost immediate recovery of the elastic limit to 
near the new yield point which will be as high as if the 
material had been allowed to rest for several days. 

Hardening by Quenching. — The hardening effect of over- 
strain is not the same as that effected by heating the metal 
to a high temperature and quickly cooling by quenching. 
This has the effect of making the metal very brittle, and 



if) 




Fig. 27. 



there is practically no yield point, the specimen breaking 
off short with practically no extension. 

Mechanical Hysteresis . — If a specimen of ductile material 
is loaded up beyond the elastic limit and the load is taken 
off slowly and the strains noted for descending loads, 
the stress-strain diagram for descending loads will be found 
not to coincide with that for the ascending load, the two 
curves forming a loop as indicated in Fig. 27 ; this, by 
analogy with magnetic hysteresis, is called a mechanical 
hysteresis loop. In experiments of this kind great care is 
necessary to eliminate errors of the instrument on the return, 
but many experimenters have found similar results well 
within the elastic limit. Very careful experiments by 
Mr. Bairstow, however, at the National Physical Labora- 
tory,! suggest that this phenomenon does not occur unless 
* Phil. Trans. Roy. Soc, 1889. t ^^id., vol. 210. 



60 



THE STRENGTH OF MATERIALS 



the " natural " elastic limit is passed. These " natural " 
elastic limits are dealt with on p. 87. 

Tensile Strength of Various Steels and Wrought 
Iron. — Fig. 28 shows typical stress-strain diagrams for various 

60 



50 



40 



30 



ZO 



10 



















/a 
































k 


1 
































/r» 


^ 






^ 








/ ^ 


















^O^ 






^ 


^ 






^y 


/i 


^^\ 








\ 





















































Extension per cent. — 
A Tool steel 

(Unannealed). 
B Crucible steel. 



HO 



30 



C Medium steel. 
D Mild steel. 
E Wrought iron. 



4o 



Fig. 28, — Stress-strain Diagrams for various Steels and 
Wrouffht Iron. 



kinds of steels and wrought iron; the strength properties 
depend to a large extent on the heat treatment and amount 
of " working " in manufacture. 



BEHAVIOUR OF MATERIALS UNDER TEST 61 

Effect of Varying Amounts of Carbon on Strength. — 
The effect of increasing the carbon in the steel is to increase 
the strength at the expense of the ductility. The following 
figures are taken from Harbord and Hall's Metallurgy of 
Steel (Griffin) for normal steels. 

Stresses in Tons per sq. in. 



Percentage of Carbon. 


Breaking Stress. 


Elastic Limit. 


•09 


21 


9-4 


•16 


29 


13-0 


•15 


33 


13-1 


•34 


35 


11-9 


•44 


41 


16-2 


•65 


54 


18-0 


•79 


57 


20^0 


•94 


62 


. 21-9 



The proportion of carbon does not have an appreciative 
effect on the value of Young's modulus, nor does tempering 
or other hardening process. 

Alloyed Steels. — The following figures give mean values 
for some examples of various alloyed steels. 



Kind of Steel. 


Tons per sq. in. 




Breaking 

stress. 


Elastic 
Limit. 


Z Elongation. 


Nickel 

[•2 % C, 3-2 % M] 


42 


27 


26 on 3 in. 


Tungsten 

[7-15 % C, -29 % Mn, -40 % W.] 

Annealed 

Unannealed 

[•46 % C, -28 % Mn, 8-33 % W.] 

Annealed 

Unannealed 


25-5 
310 

42-5 
64-0 

30-6 


18 
24 

25-5 
45-0 


39-6 on 2 in. 
33 

32-6 

2-57 


Vanadium 

[•20 % C, -27 % V, -48 % Mn] . 


25-7 

18 
49 


33-5 on 2 in. 


Chromium 
[•4 % C, 5 % Cr.] 

Annealed 

Hardened 


55 
32 


24 on 2 in. 
12 



62 



THE STRENGTH OF IVIATERIALS 



" Quality Factor." — This term has been used by some 
engineers for the result of adding the breaking stress to the 
percentage elongation. 

Stress-strain Diagrams for various Ductile Metals. 
— Fig. 28a shows tjrpical stress-strain diagrams for a number 



40 



35 



3C 



c 25 



o 



20 



15 



10 



O 











. 
















^___ 


— 


- 
























/ 














y 


/■ 














/ 














/ 


/ 














// 
















//b 




c^^ 












/ 
















/ ^ 


/ 




_D_ 












1 / 
















/ / 




^ 












//^ 
















^ 

































lO 

Extension per cent. — 

A Aluminium Bronze. 
B Hard Brass. 
C Annealed Brass. 



20 



30 



40 



D Rolled Annealed Copper. 
E Rolled Aluminium. 



Fig. 28a. — Stress-strain Diagrams for various Metals. 



of ductile metals. These must be regarded as only average 
diagrams, because these metals vary in their elastic proper- 
ties to a considerable extent, depending on the method of 
working and upon their constitution in the case of alloys. 

In most cases the early portion of the stress-strain diagram 
is never quite straight, but there is usually a clearly defined 
yield point. 



BEHAVIOUR OF MATERIALS UNDER TEST 63 



Effect of Temperature upon the Strength of Steel. 

— The effect of temperature upon the ultimate strength of 
steel is to first cause a slight diminution, then an increase 
up to about 500° F., and finally a progressive diminution in 
strength for temperatures beyond ; the elastic limit, however, 
falls progressively as the temperature increases. This is 
shown in Fig. 29, which represents the mean results of 

60 



50 



40 



^0 



20 



10 



_-^ 


_ _<^ _ 


^N 


H 


' 


^oT 











\ 


\A 








































<ao''? 






x^ 










B"--- 




"^K,^^ 












• 





2.00 4-00 600 SOO lOOO \Z0O |40C° 
Fig. 29.— Temperature Effect on Strength of Mild Steel. 

experiments on steel made at Watertown Arsenal, U.S.A., in 
1888. Stresses are in 1000 lbs. per sq. in. 

Curve A shows the variation of ultimate strength, and curve 
B the variation in elastic limit. 

For the effect of temperature upon the strength of various 
other materials the reader is referred to Johnson's Materials 
of Construction (Wiley & Sons). 

Compressive Strength of Ductile Metals. — When a 
ductile metal is tested by compression upon a short cylinder 



64 



THE STRENGTH OE MATERIALS 



(Fig. 30a) (in long specimens the failure will always take place 
by buckling ; and it is this which determines the safe stresses 
in compression members), the cylinder reaches a yield point 
which usually agrees very fairly well with that in tension and 
then bulges out almost indefinitely, as indicated in Fig. 306, 
to a slightly reduced scale, until it fails by cracking trans- 
versely. It is not alwa3^s possible to cause breaking by com- 
pression, because the flow becomes so great and the recorded 
results of ultimate crushing or compression tests therefore 
show considerable variation and are not of very great value. 
To obtain anything like a reliable result we should always 




(a) 



Fiu. 30. 




allow for the changed area in a similar manner to that 
described for obtaining the true strength in tension. 

Shear Strength of Metals. — It is not easy to obtain 
a condition of true shear in testing. Fig. 31 (a) shows the 
deformation in shearing on launching a piece out of a bar 
or plate of ductile material. The lines across the bar indicate 
lines initially parallel and the deformation is such as to 
cause tensile and compressive stresses across the rectangles 
shown in dotted lines. Alongside, in diagram {b), is shown 
a sketch of an actual shear failure of a special phosphor 
bronze taken from a paper by Mr. E. G. Izod.* The most 

* Proc. I. M. E., 1906. 



BEHAVIOUR OF MATERIALS UNDER TEST 65 

accurate method of testing for shear is by torsion u^^on thin 
tubes. 

TIMBER 

Timber is not an isotropic material, i.e. its strength 
properties are not the same in all directions, and there is 
considerable variation in the results of tests for the same 
kind of timber. This is because the strength depends upon 
the age of the timber, its dryness, the portion of the tree 
from which it has been cut, and even upon the kind of soil 
upon which it has been grown. 

The strength of timber is greatest when the weight of 



,:cr2^52S3^ 



-VV- 



-^M- 



-rs^ 



55 



V 



■//ym0i^ 



^ 



(CV.) 



iyv/. 



^ 



Fig. 31. — Shear Failures. 




moisture is about 5 per cent, of the total and decreases to 
about half this when the timber is green or very wet. 

The moisture is usually determined by taking shavings by 
boring and weighing them before and after drying in an 
oven at a temperature of about 212° E. 

In scientific tests of timber, such as Bauschinger's tests,* 
a standard moisture of about 15 per cent, is usually taken. 
Average values of strengths of various kinds are tabulated 
on p. 82. 

Tensile Strength. — The tensile strength of timber is 
very much greater when the pull is parallel to the grain 
than when it is across it. Considerable trouble is experi- 
* See Unwin's Testing of Materials (Longmans). 



66 THE STRENGTH OF MATERIALS 

encecl in gripping the specimen satisfactorily in a tension 
test on account of the tendency to shear or crush the ends. 

The stress-strain diagram for straight -grained timber for 
tension parallel to the grain is practically straight up to 
fracture. 

Compressive Strength. — In the compression of short 
cylinders or cubes of timber in a direction parallel to the 
grain, the lateral swelling causes the wood to split up into 
a number of strips or thin tubes which fail by budding, 




Fig. 32. — Compression Failure of Timber 

the line of failure usually following an inclined line as 
indicated in Eig. 32. 

Care must be taken that the ends are quite parallel, so 
that the pressure is uniformly distributed. 

When tested by crushing across the grain, the strength of 
timber is less than when the pressure is parallel to the grain. 
This strength is more one of hardness, i.e. resistance to 
penetration, and Johnson regarded the ultimate strength in 
this direction as that which gave 15 per cent, of indentation. 

Shear Strength. — The shear strength of timber is, as we 
would expect, very much less along the grain than across 



BEHAVIOUR OF MATERIALS UNDER TEST 67 

it, and in most bending tests of timber the initial failure is 

really by shear along the grain rather than by tearing of the 

fibres. 

As shown m Chap. XVI. the maximum shear stress upon a 

rectangular section of breadth h and depth d for a central 

load W is 

1-5 W -75 W 



5 = 1"5 mean shear stress = 



2,hd 



hd 



from this the shear strength of timber along the grain can 
be calculated indirectly by finding the load on a relatively 
short beam which wiU split the timber lengthwise as opposed 
to tearing the fibres. 

The results of tests by Mr. Izod * by direct shear are 
shown by the following table — 



Sheab Strength op Timber (Izod's Experiments). 
{Stresses in lbs. per sq. in.) 



Kind of 
Wood. 


Ultimate 

Tensile 

Strength. 


Ultimate Shear Strength. 


" Crippling " 

Stress across 

Grain. 


% Moisture. 


Along Grain. 


Across Grain. 


Pine . 
Oak . 
Deal . 
Teak . 


9,200 
16,000 

7,800 
9,800 


470 

890 

440 

1,000 


4,900 
5,300 
2,700 
4,000 


1,900 

1,200 

2,800 


15-7 
12-6 
10-6 
100 



The " crippling stress " is the stress at which the timber 
was found to shear through about three-fourths of its area ; 
considerable increase of load, however, was required before 
complete shear occurred. 

Bending Strength. — Tests by bending form one of the 
most satisfactory methods of testing timber. If the section 
is rectangular, of breadth b and depth d and the span is I, 
then we have for a breaking central load W as proved in 
Chap. VII. 

Breakmg stress = -^ -^ -^ = ^-^^ 
* Proc. I. M. E., 1906 (1). 



68 



THE STRENGTH OF MATERIALS 



As we indicated on p. 51 this breaking stress is some- 
times called the " modulus of rupture.' 

Young's modulus may be calculated by finding the de- 
flection S for a given safe load W bv the formula 



6 = - 



E = 



4SEI ' 

W P . 



\\P 

4:b(P .E 



111 these bending tests it is a good plan to take a standard 
>ize of beam, e.g. l" x V x 12". 




Fig. 33. — Compression Faikire of Concrete Cube. 



STONE, CONCRETE, CEMENT AND LIKE BRITTLE 

MATERIALS 

Compressive Strength. — When stone, concrete, cement 
and like materials are tested in compression in the form of 
cubes or short cylinders, fracture nearly always occurs by 
splitting in diagonal planes in the manner indicated in Fig. 33. 
This is commonly referred to as a " shear failure,'" the failure 
being attributed to the shear stresses on the diagonal planes 
at 45" to the axis. We have seen already (p. 10) that on 
such planes there is a shear stress of equal intensity to the 



BEHAVIOUR OF MATERIALS UNDER TEST 69 

compressive stress. There are, however, strong reasons for 
supposing that the fracture of brittle materials is due to 
tension and the most careful experiments on cement and 
concrete show that the shear strength is greater than the 
tensile strength (cf. p. 79). 

Tension Theory of Failure. — When a block of material 
is compressed longitudinally it swells laterally, as shown in 
dotted lines in Fig. 34a, the ratio between lateral swelling y 
and the longitudinal compression x being Poisson's ratio [r]), 
and one theory is that the [limit of compressive strength of 



ir-y 




Fig. 34. 



the material is reached when the tensile strain y reaches the 
limit of tensile strain for the material ; in the case of a block 
in a testing machine, there is nearly always a very large 
friction force F (Fig. 346) induced, which prevents the lateral 
expansion and causes the block to bulge as indicated in 
dotted lines, thus causing resultant stresses R in a diagonal 
direction, which cause the apparent shear fracture and make 
the specimen to appear stronger than it really is. It has 
been kno^vn for many years that the measured compressive 
strength of blocks depends upon the material placed between 
the press head and the block. The following figures quoted 
from Unwin's Testing of Materials are of interest in this 
connection — 



70 



THE STRENGTH OF IVL^TERIALS 



Crushing Load 
Material. in tons on 
4 in. cubes. 


Material between cu1)e and preas head of i 
Testing Machine. 


Portland stone 

1 


57-7 
52-6 
450 
33-5 


Two millboards | 
One lead plate | 
One lead plate J in. smaller all round ! 
Three lead plates 


Yorkshire grit 


79-7 
800 

56-2 
35-9 


1 
Two millboards 
Cemented between two strong iron 

plates with plaster of Paris 
One lead plate 
Three lead plates 



The lead plates were '085 in. thick in each case, and the 
fracture was longitudinal, as in Fig. 34a, in each case with 
lead plates, and diagonal with the millboards. Professor 
Unwin, in commenting upon this, says that the "lead falsifies 
the result of the experiment," but we do not see why he 
should not consider the lead as giving the more correct result 
and the millboard figures as being false. 

Professor Perry apparently takes the latter view, for he says 
in Applied Mechanics (Cassell) : " There is much published 
information on the fracture by compression of blocks of 
stone, cement and bricks. In almost every case care is taken 
in loading the usually short specimens that friction at the 
ends shall prevent the material sweUing laterally. When 
sheet lead is inserted at the ends, it gives a small amount 
of lateral freedom, and in everj'- case the breaking load is 
lessened by its use, and therefore it is said to be wrong to 
use lead. I consider all this published information to be 
nearly valueless, except that there is some probability that 
half the usually published ultimate compressive strength 
for a cube is the true resistance to compression in the 
material.*' 

There is, of course, in the case of a pure compression, a 
shear stress across a diagonal plane, and for materials like 
mild steel, in which the shear strength is less than the com- 
pressive strength, tliis shear stress probably causes the ultimate 
failure, and thus determines the compressive strength. 



lateral strain 


= 2/ 


= 7] X 


= 


>7K 

e; 


tensile stress 


= Ut 


= ^ • 


u. 


• E, 







BEHAVIOUR OF MATERIALS UNDER TEST 71 

Batio between Tensile and Compressive Strength for Con- 
crete. — The consideration of the transverse strain enables us 
, to calculate the ratio between the tensile and compressive 
strength of the material upon this theory. 

Let Uf = ultimate tensile strength of the material. 

u, = ultimate compressive strength of the material. 
E, = Young's modulus in tension at failure. 
E, = Young's modulus in compression at failure. 
r) = Poisson's ratio. 

Then, compressive strain = x = ^^- 



(1) 
Ratio of tensile to compressive strength 

-u:~~e: ^^^ 

Now this ratio, according to different authorities, varies 
from one-eighth to one-twelfth, according to the usual method 
of determining compressive stress, and depends on the age 
of the concrete, the higher value occurring usually at ages of 
three months and more, rj for concrete is not fully known, 
and in the absence of further information we will assign to it 
the value J, which is the theoretical value for a perfectly 
elastic solid. 

E< also has not been very fully determined, but Hatt, for a 

1:2:4 mixture gives E< = 2*1 x 10" lb. per sq. in., which is 

approximately equal to the value usually accepted for E,. 

E, 
For a rough consideration, therefore, we will take -^p = 1. 

This would give -'-=■• 
Uc 4 

If, as Professor Perry suggests, the actual compressive 
strength is about one-half that usually published, the above 



72 THE STRENGTH OF MATERIALS 

u 1 
figure compared with published figures gives ~ = q, which 

lie O 

bears comparison with the figures usually given, agreeing 
with the lower limit. 

All of these points are of very great importance, and it 
would be of great value if ver}^ careful experiments were 
made to determine ?;, E^ and E, for the same mixture at the 
same age, and to see how nearly true is the suggestion that 
the compressive strength is given in terms of the tensile 
strength by the above formula. 

Another consideration which enters into the problem is 
that tensile strengths as determined by the usual briquettes 
are somewhat less than the actual tensile strengths, the 
discrepancy being due to the variation in the distribution of 
the tensile stress across the specimen (see p. 78). According 
to various authorities the actual tensile strength is 1"5 to 
1*75 the mean strength, and allowance for this would bring 

u 11 

the value of — from ^ ^ to ^ . , which agrees very well with 
u, 12 14 ° "^ 

experimental values. 

We have already dealt with Navier's theory for this 
problem (p. 45). 

Effect of Relative Height and Breadth of Com- 
pression Specimens. — Very careful experiments in 1876 by 
Professor Bauschinger upon sandstone prisms have shown 
that the compressive strength of sandstone prisms decreases 
slightly when the relative height to breadth is greater 
than for a cube and increases when the relative height is 
less. 

Fig. 35 shows a curve expressing the results of Bauschinger's 
tests and is a modified form of a similar curve given by 
Professor Johnson,* and tends to support both the Navier 
theory and the transverse tension theor}'' that we have just 
given, because in one case we should have that the cube and 
more dumpy specimens prevent the rupture along the line given 

* The Materials of Construction (Wiley & Sons), 



BEHAVIOUR OF MATERIALS UNDER TEST 73 



by the theory, and in the other case the effect of the friction in 
preventing the natural transverse swelling is greater with a 
dumpy section than with a comparatively tall one. 

I -5 




O I 

Eatio 5eight 
Breadth 

Fig. .35. — Effect of Height upon Strength of Sandstone Blocks in 

Compression. 

Bauschinger recommended the following formula to repre- 
sent the results of these tests — 



u. 



4A/ _^.V'k 

p \ h 



where u, = ultimate crushing stress 
A = area of cross-section 
p = perimeter of cross-section 
h = height of specimen 
a, b = constants. 

Strength of Cube with Chamfered Edges. — Fig. 36 
shows the results of Bauschinger' s tests upon chamfered 
specimens, the part of the curves in full lines representing 
the range over which the actual experiments were carried. 



74 



THE STRENGTH OF MATERIALS 



The curve marked A is for comparison of the strength of 
the chamfered block with a cub^ of the same size as the large 
area, and the curve marked B is for comparison of the 
strength of the chamfered block with that of a cube of the 
same size as the small^area. 

L 



o 



•0 


































/ 






\ 










/ 




•8 




A 






— ' 


7- 






•f> 






V 


X 


z^ 














A 


< 


i 


^- 


^^_ 




•A- 


/ 


Va 
















/ 












^ 






y^ 


^ / 




/ \ 


yA 




.? 



























































•4 <A 



u 13 



-~~~-\Otr 



O 



^ 



"O 



•4 6 B 

Compressed Surface -^ Total Area 
-Compression Strength of Cube with Chamfered Edges. 



"L 

Fm. 36.- 

Stress-Strain Diagram for Portland Cement and 
Concrete in Compression. — The kind of concrete which 
we will consider is composed of a mixture of Portland cement, 
sand and broken stone or brick, gravel or like material which 
is called "aggregate." 



BEHAVIOUR OF MATEKIALS UNDER TEST 75 

The composition is usually referred to in the ratio of 
volumes of cement : sand : aggregate, i.e. a 1:2:4; concrete 
is one composed of 1 part cement, 2 parts sand and 4 parts 
aggregate. 

The stress-strain diagram for concrete in compression is 
never quite straight so that there is no elastic limit, the 
exact curve depending on the composition and on the time 
after setting. 

The curve shown in Fig. 37 is almost exactly a parabola. 



20Q& 



1500 






i 



in 



500 




•0004- 



0006 



oooe 



Strain joer In. 
Fig. 37. — Stress-strain Diagram for Concrete in Compression. 

This curve is for a 1:3:6 concrete, 90 days old, which was 
tested by Mr. R. H. Slocum, of the University of Illinois. 
Some authorities assume that the curve is a parabola, but 
in practice it is seldom that the curve comes so near to a 
parabola as the above. The stress-strain curve is, however, 
nearly always of a similar shape, the strains increasing more 
quickly than the stresses. It is extremely important to 
remember that with cement and concrete the relations between 
stress and strain vary largely with the quality and pro- 
portions of ingredients, and cannot be taken as almost 
constant as in the case of steel. In tension a somewhat 



76 



THE STRENGTH OF :\IATERIALS 



similar curve is obtained, but as ceraent and concrete are 
practically never used in tension, much less work has been 
done on its tensile properties. 

Young's Modulus for Concrete. — In a material like 
concrete Young's modulus E is not constant, so that ^\e 
must give the stress at which the ratio is taken if it is to 
have any real value. 

The initial value of E is obtained b}' drawing a tangent to 
the curve at the origin as indicated in the figure. 




Fig. 38. 



We then have initial E 
Similarly final E 



1200 
•0004 
1800 
•0012 



= 3 X 10^ lbs. per sq. in. 
= To X 10^ lbs. per sq. in. 



The usual value of E taken in reinforced concrete calcu- 
lations is 2 X 10^ lbs. per sq. in. 

Effect of Composition and Age upon the Compressive 
Strength of Concrete. — The compressive strength of con- 
crete is roughly proportional to the i^roportion of the cement 
in the mortar. Fig. 38 shows a diagram plotted from the 
results of experiments by ^Mr. G. W. Rafter, of Xew York. 



BEHAVIOUR OF MATERIALS UNDER TEST 77 

Clean, pure silica sand and Portland cement were used, and 
the aggregate consisted of sandstone broken so as to pass 
through a 2 -inch ring, containing 37 per cent, of voids when 
rammed. 

The compressive strength increases with age, and Fig. 39 



v^- 4CQO 

I 

Co 

^i i>oocK 

- C3 

"^^ 2000 

\ 

CO 

.S 1000 

I 

I 




f 2 3 ^ 

Fig. 39. 



shows on a diagram the results of experiments made at the 
Watertown Arsenal, U.S.A., in 1899. 

Curve A is for a mixture of one part of cement, two j)arts 
of sand, four parts of aggregate ; and curve B is for a mixtui-e 
of 1:3:6. The figures given are for the same brand of 
cement. 

Tensile Strength of Portland Cement and Concrete. 
•^The tensile strength of concrete is about -^^ of its com- 
pressive strength, but it is not usual to allow for any 
tension in the concrete in practice. 

The standard method of testing the strength of neat cement 
is, however, by tension, so that the tensile strength is of 



78 



THE STRENGTH OF MATERIALS 



considerable importance. (For method and apparatus for 
testing, see Chap. XIV.) 

The British standard specification requires the following 
strength of Portland cement. 

Briquettes 1 sq. in. in section (see Fig. 190) must develop 
at least the following strength — 
Neat cement. 

After 7 days (1 in moist air, 6 in water) . . 400 lbs. 
,, 28 ,, ,, ,, 27 ,, . . oOO ,, 




Fig. 40. 



The increase from 7 to 28 days shall be at least — 



25 % when 7 -day test gives between 400 lbs. and 450 lbs. 



20 0/ 



450 „ 500 „ 
500 ,, 550 ,, 
550 „ 600 „ 
600 lbs. or upwards. 



1^0/ 

J-" /O ?J J 5 5> 5 5 

^^ /O 5 5 5 5 5 5 5 3 

K 0/ 

<-' /O 5 5 3 5 3 J 3 3 

One 'part cement and three parts sand. 
After 7 days (1 in moist air, .6 in water) . . 150 lbs. 

„ 28 „ „ „ 27 „ . . 250 „ 

Increase between 7 and 28 days must be 20 % for stresses 

200-250 and 5 % less for each 50 lbs. increase, 5 % 

being the minimum. 

Variation of Stress in Briquette. — It can be shown that 

the stress is not quite constant over the briquette, but varies 

somewhat as indicated in Fig. 40; this means that the test 

strengths are always a little less than their actual values.* 

Strength of Concrete in Shear. — Early experimenters 

found the shear strength of concrete to be from '12 to '2 of 

* See also Chap. XIV. 



BEHAVIOUR OF MATERIALS UNDER TEST 79 

its compressive strength, but recent experiments suggest that 
the shear strength is considerably greater than this and 
depends upon the kind of aggregate (gravel, broken stone, 
broken brick, etc.) used. The most exhaustive investigation 
is that by Professor A. N. Talbot, who has also carried out 
many valuable experiments on the strength of reinforced 
concrete beams. The results of these experiments were pub- 
lished in a Bulletin of the Engineering Department of the 
University of Illinois. 

Two different methods of testing were used; in the first 
the shear strength was obtained by punching a* hole in a 
concrete plate, and in the second by means of concrete beam 
with the ends fixed. There was considerable variation in 
the results, as will be seen from the following summary taken 
from Engineering of June 6, 1907 — • 

SuMMAHY OF Shear Tests. (Professor Talbot.) 



rorm of 
Specimen. 



Plain plate 



Recessed 
block 



Reinforced i 
recessed ' 
block 

Restrained 
beam 



M 

o 

C 

o 1 

o 






1 

1-3-6 1 


1-3-6 : 


1-3-6 


1-3-6 ! 


1-2-4 


1-3-6 


1-3-6 


1-3-6 


1-3-6 


i 1-3-6 


: 1-2-4 ! 


! 1-3-6 


1-3-6 


1-3-6 j 


. 1-2-4 1 


1-3-6 ' 


1-3-6 


1-2-4 



Method of 
storing. 



Strength. 



Ratio of 

Shear to 

Compression. 



Air 

Water 

Damp sand 

Do. 

Do. 

Air 
Water 

Do. 
Damp sand 

Do. 

Do. 

Air 
Damp sand 

Do. 

Do. 

Do. 

Do. 

Do. 



g \ Shear. 



Compression. 



Cube. ^Jl^"- 
I der. 





lbs. per 




sq. m. 


9 


679 


7 


729 


4 


905 


1 


968 


5 


1193 


17 


796 


6 


692* 


5 


879 


4 


1141 


1 


910 


5 


1257 


4 


1051 


4 


1821 


1 


1555 


5 2145 


4 


1313 


1 


1020 


6 


1418 



lbs. per 
sq. in. 

1230 
1230 

2428 
1721 
3210 
1230 
1230 
1230 
2428 
1721 
3210 
1230 
2428 
1721 
3210 
2428 
1721 
3210 



lbs. per 
sq. in. 



1322 
1160 
2430 



1322 
I 1160 
j 2430 

1322 
': 1160 

2430 
: 1322 
i 1160 
I 2430 



C-be. I CJ^- 



0-55 
0-59 
0-37 
0-56 
0-37 
0-65 
0-56 
0-71 
0-47 
0-53 
0-39 
0-86 
0-75 
0-90 
0-67 
0-54 
0-59 
0-44 



0-68 
0-83 
0-49 



0-86 
0-78 
0-52 

1-38 
1-39 

0-88 
1-00 
0-88 
0-58 



* Specimens injured in removing the forms. 



80 THE STRENGTH OF MATERIALS 

The conclusions to which these Illinois exj)eriments lead 
are that the resistance of concrete to shear is dej)endent on 
the strength of the stone used, as well as on the strength of 
the mortar ; and in the richer mixtures the stone appears to 
exercise the greater influence. With hard limestone and 
1-3-6 concrete sixty days old the shearing strength may be 
expected to reach 1100 lbs. per sq. in.; and with 1-2-4- 
mixture 1300 lbs. per sq. in. There is reason to believe that 
if tests can be made with the load applied evenly over the 
shearmg section, so as to obtain the true resistance to simple 
shear, the results will be found to be higher than those already 
obtained. 

An important point brought out by Professor Talbot's 
investigations was the influence which variations in the 
constitution of the concrete have on the shearing strength. 
The compressive strength of concrete is largely affected by 
the strength of the cement, but the shearing strength is 
influenced more by the strength of the aggregate. Eor this 
reason it does not seem well to express the shearing strength 
in terms of the compressive strength. The method has the 
advantage, however, that an idea is gained of their relative 
action. 

If, as indicated by these experiments, the shear strength of 
concrete is greater than the adhesion between concrete and 
steel, then there is an advantage over plain bars for reinforce- 
ment in those bars such as the twisted or indented bars 
which cannot be withdrawn from the concrete without 
shearing it. 

Adhesion between Concrete and Steel. — It is absolutely 
necessary in a reinforced concrete structure that there shall 
be a good bond between the concrete and the steel, for the 
latter will bear its share of the stress only so long as there is 
no relative movement between the steel and the concrete. 
If a concrete beam were cast with holes throughout its length 
on the lower side and steel rods were inserted loosely into 
these holes, the strength of the beam would be practically 



BEHAVIOUR OF MATERIALS UNDER TEST 81 

no greater with the rods than without, because relative move- 
ment between the steel and concrete would be possible. A 
concrete beam with the reinforcing bars loose is like a plate 
girder without an}^ rivets. The adhesive strength for plain 
bars can be found as follows : Let O be the perimeter of the 
bar and I its length ; then if / is the safe adhesive stress, the 
adhesive force T that can be carried is given by 

F = Olf 

f can be found experimentally by embedding a rod in concrete 
and finding the force necessary to pull it out and dividing 
the resulting stress by the factor of safety (usually taken as 
about 6). Most authorities take a safe adhesive stress of 
60 lbs. per sq. in. 

Numerical Example. — Find the length in relation to the 
diameter of a round bar that must he embedded in concrete in 
order that the tensile stress of 16,000 lbs. per sq. in. will be reached 
as soon as the safe adhesion stress of 60 lbs. per sq. in. 

Let d = diameter of rod in inches. 

Let I = length of rod in inches. 

Then load to reach safe tensile stress 

== P = stress X area 

= 16000 X ^ 
4 

Load to reach safe adhesive stress 

= P = stress X length x perimeter 
^ QO X I X 7rd 

Trd^ 

If these are equal, 601 x -n-d ^ 16000 x . 

16000 wd^ 
~ 60 X 4 ^7rd 
= 61 d nearly 

. • . the bar must be embedded for a length equal to 67 
diameters. ^ 



82 



THE STRENGTH OF IVUTERIALS 



Strength of Tn^reER (Normal Values) 

{Stresses in thousarhds of 'pounds per sq. in.) 



• 




Ultimate Strength. 




! 


1 


Kind of Timber. 










Young's 
Modulus 
(millions 
of lbs. per 
sq. in.). 


Weight 
in lbs. , 
per 1 
cu.ft. 


Tensions 
along 
Grain. 


Com- 
pression 
along 
Grain. 


Shear Shear 
across along 
Grain. Grain. 


Bend- 
ing. 


Ash .... 


8-14 


6-8 


2-4 


•4--7 


10-12 


1-6 


50 1 


Beech . 


9-18 


4-6 


— 


— 


8-10 


1-3 


44 , 


Dantzig Fir 


6-9 


3-5 


2-7 


•4 


4500 


o 


33 ! 


Elm . . . 


6-12 


5-8 


3-5 


•6--9 


6-8 


1-6 


34 1 


Oak. . . . 


9-15 


4-5-9 


3-5-5 


•6--9 


7-10 


1-7 


58 


Pitch Pine . . 


8-12 


4-6 


4-9 


•4 


7-5 


1-5 


42 


Red Pine . . 


6-9 


4-6 


3 


•4 


4r-6 


1-2 


27 


Teak . . . 


12-14 


12-14 


4 


1 


12-16 


2-4 


49 


Yellow Pine . 


6-9 


4-6 


3-5 


•4 


4-8 


1-7 


32 

1 



Normal Crushing Strength of Cement, Stones, etc. 



Material. 



Brick (London stock) . 

,, (Staffordshire blue) 
Brickwork in Cement . 
Cinder Concrete (1 : 2 : 4) 

Granite 

Gravel Concrete 1:2:4 
1:3:6 
Portland Cement . 
Portland Stone , 
Sandstone .... 
Slate 



Weight in lbs. 
per cu. ft. 



Ultimate Crushing Strength. 
(Thousands of lbs. per sq.in.) 



115 

140 

100-150 

97 

170 

120 

130 
90 

145 
135-145 

175 



2-5 

7 

1-25-2-5 

1-8 after 28 days 

12-20 
2-4 after 28 days 

1*8 ,f ,5 

7 

5 

5-10 

10 



BEHAVIOUR OF MATERIALS UNDER TEST 



83 






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CHAPTER III 

REPETITION OF STRESSES : WORKING STRESSES 

Repetition or Variation of Stresses. — In the design 
of machines and structures, we very often have to deal with 
cases in which the stresses vary in amount from one time to 
another ; such cases occur in nearly every machine part sub- 
jected to rotary and reciprocatory movements and in struc- 
tures which have to resist wind-pressures and rolling loads. 
In recent years, a large amount of investigation has been 
carried out on the strength of materials which are subjected 
to alternating stresses. The stress required to cause rupture 
in a material which is gradually increasingly stressed is called 
the static breaking stress, and is the stress obtained in the 
ordinary testing machines. 

Fairbairn discovered in connection with some tests on 
wrought-iron girders, that a girder can be ruptured by 
repeatedly applying a load equal to about one-half of the 
static breaking load. 

The first exhaustive investigation on the subject was 
conducted by Wohler on behalf of the Prussian Ministry 
of Commerce, and was published in 1870. Wohler's experi- 
ments extended over a period of twelve years, and had results 
which at the time were very startling, and the importance of 
which has only in comparatively recent years been appreciated 
by engineers. 

The general result of these and subsequent experiments 

is to show that the stress necessary to rupture a material 

84 



REPETITION OF STRESSES 



85 



when such stress is repeated a very large number of times 
is considerably less than the static stress. 

In Wohler's experiments, which were carried out in tension, 
bending and torsion, some of the variations were from zero 
to a maximum in tension or compression and some were 
for a complete reversal of stress. 

In one form of Wohler's apparatus for testing by reversal 
of stress in bending, the specimen was in the form of a 
projecting beam or cantilever A (Fig. 41) clamped at the 
end of a shaft E, mounted between bearings B. The shaft 
was rotated by means of a belt surrounding a pulley C, and 



5 




c 




B 




A-^ 








_ . 








E 















L 




Fig. 41. — ^Wohler's Experiments. 

the specimen was of circular section and loaded at the end by 
a spring D, and in the rotation the compression and tension 
sides changed places gradually, thus giving a gradual reversal 
of stress. To balance the forces on the machine, a specimen 
was mounted at each end of the shaft. 

In another form of apparatus a beam was mounted upon 
knife edges to one of which a spring was connected by levers. 
The load was applied in the centre by a spring rod which 
was lifted periodically by a crank upon a rotating shaft, thus 
gradually applying the load and taking it off again. 

Full accounts of the experiments will be found in Unwin's 
Testing of the Materials of Construction. We will take some 
examples of his results : — 



86 THE STRENGTH OF MATERIALS 

For Krupp's Axle Steel — 

Statical breaking stress = 52 tons per sq. in. 
Breaking stress from zero to maximum = 26" 5 ,, ,, ,, 
„ ,, for reversed stresses = 14*05 ,, ,, „ 

For Wrought-Iron — 

Statical breaking stress = 22*8 tons per. sq. in. 
Breaking stress from zero to maximum = 1525 ,, ,, ,, 
„ ,, for reversed stresses = 8*6 ,, ,, ,, 

In the first case the range of stress is in one case 26* 5 and in 
the case of reversal is — 14'05 to + 1405, i. e. 28*1, whereas 
the corresponding figures in the second case are 15' 25 and 
17-2. 

Sir Benjamin Baker carried out similar experiments in this 
country and obtained similar results. 

For mild steel of static strength from 26' 8 to 28* 6 tons per 
square inch, he obtained a breaking stress of 11" 6 tons per 
square inch for reversal of stress. 

Bauschinger carried out a large number qf experiments on 
the same lines as those of Wohler, and extended them to a 
larger number of materials. 

For Bessemer Steel his results were — 

Static breaking stress = 28*6 tons per sq. in. 
Breaking stress from zero to maximum = 15" 7 ,, ,, ,, 
,, ,, for reversal stresses = 8*55 ,, ,, ,, 

With regard to these breaking stresses for variations of 
stress, it should be remembered that these are the least 
stresses for which the specimen would break after a very 
large number of repetitions. 

In carrying out tests of this kind a number of specimens 
are taken, and the range or amount of variation of stress is 
altered for different specimens or sets of specimens, and when 
the range comes below a certain value the specimen will not 
break within the time over which the experiment lasts. The 
results are expressed on a diagram in which the range of 
stress is plotted against the number of repetitions required to 



REPETITION OF STRESSES 



87 



cause fracture; or, in the case of variations from zero to a 
maximum or of complete reversal of stress, the limit of stress 
is plotted against the number of repetitions. Such a curve 
is shown in Fig. 42. From such curves the apparent stress, 
at which an infinite number of repetitions could be made 
without fracture, is obtained, and this is taken as the least 
breaking stress. The word "apparent" is used because no 
record appears to exist of a number of repetitions more than 
about fifty millions, and it has been suggested that perhaps 



^ 



'$5 



in "^ 






Stal'ic 3tnsss 




L/nuT for Conn^le^^e Re.{^ersa 



20 



^O 



QO 



SO 



I00< 



f^GboTit/ons (^Hunc/reJ thousanclsj 
Fig. 42. — Repetitions of Stress. 

lower stresses still would be obtained if the repetitions were 
extended still more. 

Bauschinger suggested that there was some relation between 
the range of stress which a material would stand and the 
elastic limit. This elastic limit was what he called the 
" natural elastic limit," i.e. that obtained after the material 
has been subjected to a few variations of stress. We thus get 
the theory of natural elastic limits, which states that the range 
of repetition of stress which a material can resist indefinitely 
without failure is the range between the natural elastic limits 
in tension and compression. 



88 THE STRENGTH OF MATERIALS 

Dr. Stanton and Mr. Bairstow published in Vol. CLXVI. of 
Proc. Inst. C.E. an important paper on the subject, giving 
the results of experiments conducted at the National Physical 
Laboratory. 

They used a machine in which the specimen formed part of 
the piston-rod in a steam-engine mechanism; the specimen 
thus was subjected to reversals of direct stress, and a 
variation in the limiting stresses was obtained by varying 
the relative dimensions of the mechanism. 

This research had some important results, the principal ones 
of which are — 

{a) An alteration of the rate of repetition from 60 to 800 per 
minute has no marked effect on the results obtained. 

{h) The range of stress which moderately high-carbon steels 
can stand is comparatively greater than that for 
low-carbon steel and wrought iron. This confirms 
Wohler's opinion, and is contrary to the common 
idea that a comparatively brittle material can with- 
stand less variation than a ductile one. 

(c) The limiting stress which iron and steel can bear depends 
on the range of stress, and is almost independent of 
the actual values between the limits i and f. This 
means that a stress variation from, say, 7 tons per 
square inch in tension to one of 5 tons per square 
inch in compression has the same effect as one from 
6 tons per square inch in tension to 6 tons per square 
inch in compression. 

Although the authors agree that more work must be done 
before a definite statement can be made, their experiments go 
to support Bauschinger's theory as to the elastic limits. 

Effect of Rate of Repetition upon Results. — There 
seems to be some unexplained difference in the results of 
experiments upon the effect of speed upon the results. 
Wohler's experiments were at 60 repetitions per minute, as 
indicated above. Stanton and Bairstow found no appreciable 



EEPETITION OF STRESSES 89 

effect of increasing to 800 per minute; Reynolds and Smith,* 
however, employing the machine described in Chap. XIV., 
found that there is a progressive diminution in the resistance 
against repetition for repetitions of 1300 per minute and 
upwards ; Eden, Rose and Cunningham, f however, experi- 
menting with short rotating beams with a uniform bending 
moment over an appreciable length, found no such effect for 
speeds of 1300 per minute. 

The following summary of results, pp. 94, 95, taken from 
the last-mentioned paper, gives a clear idea of the results of 
the various experiments on the subject. The only explanation 
of these contrary results appears to be in the difference in 
the design of the machines ; at the high speeds it is possible 
that some secondary influence had a marked effect upon 
the results. 

The Fatigue of Metals. — The phenomena described 
above are often referred to as the "fatigue of metals"; 
the suggestion being that the stress causes a change in the 
molecular structure of the metal and that the metal gets 
fatigued after a time and so breaks down under a smaller 
load. The bulk of the evidence, however, appears to be 
against that view and in favour of the theory that ultimate 
failure will occur only if the elastic limit is exceeded and 
thus the effects of overstrain become accumulative. 

Specimens cut out of pieces that have been fractured by 
repetition of stress do not exhibit any weakening that the 
fatigue idea suggests. 

The subject is still full of difficulties from the point of 
view of a satisfactory explanation of the results. For instance, 
the effect of overstrain is to cause the material to become 
brittle, and yet the more brittle kinds of steel (the high 
carbon steels) show less effect than the mild steel. 

One explanation, called Foster's Theory, is that the 
mechanical hysteresis (p. 59) causes a very small permanent 

* Phil. Trans. Roy. Soc, 1902. 
t Proc. I.M.E., 1911. 



90 



THE STREXGTH OF I\L\TERIALS 



strain at each repetition and that the effect of these permanent 
strains is cumulative so that ultimately the permanent strain 
becomes sufficient to cause failure. 

The appearance of the fracture in these experiments is 
always different from that for ordinary static tensile tests; 
that for mild steel being more like a hard steel. This is 
probably due to the effect of overstrain upon the properties 
of the metal. 




c : c *■ 

Carb en Feyoenta q e 



Fig, 43. — Repetitions of Stress — Abrupt Change of Section. 

Effect of Sudden Change of Section upon Results.— 
The effects of sudden change of section have been investigated 
by Dr. Stanton and ^Mr. Bairstow,* who obtained the results 
shown in Fig. 43, in which the limiting stresses for failiure by 
repeated loading are plotted against the carbon percentage. 
The results may be summarised as follows. 

1. The resistance of a screw-cut specimen varied from 
67 to 70 per cent, of the maximum resistance of the corre- 
spondmg material, the fracture always taking place at the 
end of the thread. 

2. The resistance of a specimen having a moderately 

* See Engineering, April 19, 1907. 



REPETITION OF STRESSES 



91 



rapid change of section varied from 65 to 72 per cent, of the 
maximum resistance of the corresponding material. 

3. The resistance of a specimen having a sudden change of 
section varied from 47 to 52 per cent, of the maximum resist- 
ance of the corresponding material. In this form of specimen 
the low carbon material appears to realise a larger percentage 
of the maximum resistance than the higher carbon materials ; 
but it is worthy of notice that even under conditions which 
are commonly supposed to be the most fatal to high carbon 
steels — ^^e. a sudden change of section — the actual resistance 
of the 0'4 and 0*6 carbon steels is approximately 40 per cent, 
greater than that of the iron. 




Fig. 4:3a. — Repetitions of Stress — Abrupt Change of Section. 



The above resistances are, of course, estimated per unit of 
area, so that in calculating the strength of a screwed rod 
under alternating stress it will be further necessary to take 
into account the area at the bottom of the threads, so that 
the total reduction in resistance may well be more than 50 
per cent, of its maximum value. 

In the case of screw-threads there is a further possible 
source of weakness due to faulty machining in the cutting of 
the screw. If the bottoms of the threads are not properly 
curved, but left with a sharp angle, there can be no doubt 
that risks of the development of a crack are very considerably 
increased. It seems quite probable that failures of steam- 
engine crosshead bolts, which have broken under* very low 
ranges of stress, may be due to this cause. 



92 THE STRENGTH OF MATERIALS 

Cast Iron. — Ver}^ little work appears to have been done 
on repetitions of stress for cast iron, but from a small number 
of experiments by the author in reversal by bending the 
same general result was obtained, the limiting breaking 
stress in this case being nearly one-quarter of the static 
stress. 

Equivalent Stress Formulae. — Straight line Formula. 
— If /,. is the greatest stress that can be applied for an 
indefinite period for a range of stress r, and f, is the static 
breaking stress of the material, the results of repeated 
load experiments can be expressed approximately by the 
relation 

t, = f.-r 

For a variation from zero to /„ i. e. r = /^ ; this gives /,, = ~ 
For a reversal /, to — /„ i.e. r = 2/,,; this gives /, = -^ 

Unwin's Formula. — Unwin has given a formula from 
which the equivalent static stress for a given range of stress 
can be found. This formula gives, when plotted, a curve 
sometimes known as Gerher's 'parabola. 

The formula is 

fe = 2' + ^ ifT^^^^r'f]) 

where n is a constant depending on the nature of the material . 
For mild steel we may take n = 1'5. 
Now if the variation is from zero to /,., then r =^ f. 

Solving this equation we get fe^'^fs- For complete 
reversal r = f, — {— f^) = 2fc 

or, /. = J /s. 
Relation between Repetition of Stress and Sudden 
Loading. — The similarity between the results of experiments 
on the variation of stresses and the reasoning given on p. 34 



REPETITION OF STRESSES 93 

with regard to sudden loading has led many authorities to 
think that Wohler's experiments were really experiments on 
sudden loading. The alternative point of view is that the 
two questions are distinct, and that therefore separate allow- 
ance should be made for each in the design of machines and 
structures. 

One of the first difficulties to overcome in reconciling the 
questions is that strain is not proportional to stress beyond 
the elastic limit, and that, therefore, beyond this point twice 
the strain would not cause twice the stress (see Fig. 2). 
There is, however, the fact that if a material is strained 
beyond the yield point, the yield point will be found to have 
been raised on a subsequent testing ; therefore, if this action 
goes on indefinite^ with each repetition of stress, the yield 
point will ultimately become so high that the dynamic 
argument will apply up to the breaking point. 

Although there are still many points which require to be 
decided in this controversy^, for practical reasons we prefer 
to allow for one or the other, but not both, in design. The 
reason for this is as follows : Suppose that the safe working 
stress for mild steel for a constant and gradual load is 7*5 
tons per square inch. Then, on the dynamic theory the safe 
stress for a reversing and sudden load is one-third of this, 
i. e. 2*5 tons per square inch. If we now make a separate 
allowance for the repetition of stresses, our working stress 

would be vv X 2*5, or "8 ton per square inch. As there is no 

question of impact in this, this seems an absurdly low working 
stress, and experience shows that it is not necessary to make 
the allowance for both points of view. 

WORKING STRESSES 

The Conflict between Theory and Practice. — An 

engineer has been tersely described by a somewhat char- 
acteristic American as " a man who can do for one dollar 
what a fool can do for two." Although from an aesthetic 



94 



THE STRENGTH OF MATERIALS 



SUMMARY OF RESULTS OF EXPERIMENTS 

Quoted from Eden, Rose and Cunningham 

Note. — The values for ** Range of Stress " causing fracture after 10® 
alternations of stress in this Table are copied from the published accounts 
of the experiments of Turner, Reynolds and Smith, and of Stanton and 
Bairstow. The corresponding figures for the tests of Wohler, Baker, and 



Experi- 
menter. 



Wohler* 
Baker* 



Rogerst 



Eden, Rose 

and 
Cunning- 
ham t 



Turner§ 



Stanton 

and 
Bairstow|| 



Reynolds 
and 

Smithy 



Type of 
Endurance Test. 



Material. 



Rotating 
Cantilever 

Rotating 
Cantilever 

Rotating 
Cantilever 



Rotating Beam 

Uniform 

Bending Moment 

Eixed cantilever 

rotating deflection 

Reciprocating 

Weight 

dhect tension and 

compression 

tension 
ratio 



j Phoenix Iron 
Homogeneous Iron 
Vickers' Steel Axle.s 
I Firth's Tool Steel 

Soft Steel 
I Fine Drift Steel 
|Ste«lC(0-32%C.) 
I as rolled 

Steel C annealed 

J-in. bright-drawn 
Wrought-Iron bar 

; ^-in. bright-drawn 
MHd-SteelrodA 

I MUd Steel 
Nickel Steel 



compression 
= 1-4 



Reciprocating 

Weight 

direct tension and 

compression 

tension 

ratio compression 

= 115 



Wrought Iron No. 2 
Piston-rod Steel 



Mild Steel annealed 



Cast Steel annealed 



Tensile Test Figures. 



Tenacity. 

Tons per 

square inch. 



21-3 

28-1 

27-7 
55 

27-7 
54 



29-3 
26-5 



33-8 



35-7 

27-3 
48-0 



250 
43-8 



25-8 



48-0 



Limit of 

Elasticity. 

Tons per 

square inch. 



16-7 
71 



26-5 

250 

l8^ 
361 



13-4 
19-6 



* The Testing of Materials of Construction. Unwin. 
•\ " Heat Treatment and Fatigue of Iron and Steel." Rogers. Journal 
of Iron and Steel Institute, No. 1, 1905. 

% For other metals see Proc. Univ. of Durham Phil. Soc, vol. iii. p. 

251. 

§ " The Strength of Steels in Compound Stress, and Endurance under 



REPETITION OF STRESSES 



95 



UPON THE REPETITION OF STRESS. 

on the endurance of metals {Proc. I, M. E., 1911). 

Rogers have been estimated from stress -revolutions diagrams plotted from 
the various published results. 

The tenacity figures for the Reynolds and Smith tests are for short 

















Endurance Test Figures. 






Range of Stress 




i 




Speed. 


causing frac- 








Alternations 


tiue after 10'' 


Eatio 


Ratio 




of Stress 


alternations. 


Eange of Stress 


Range of Stress 




per minute. 


Tons per 
square inch. 


Tenacity 


Limit of Elasticity 


60-80 


22-8 


1-07 






60-80 


24-5 


0-87 






60-80 


24 


0-87 







60-80 


30-9 


0-56 


— 


50-60 


25 


0-90 




50-60 


27-5 


0-51 


— 


400 


32 


1-09 


1-92 




400 
{ 250 


27-7 


105 


3-9 










\ 620 


34-6 


102 


1-3 




I 1,300 










r 300 

\ 600 

[ 1,300 

250 










39 


1-09 


1-56 












35-6 


1-3 


1-9 




250 


52-5 


11 


1-46 




800 


19-2 


0-75 


1-43 




800 


28-3 


0-65 


1-44 




' 1,337 


20-9 


0-81 







1,428 


20-1 


0-78 







1,516 


19-2 


0-75 







1,656 


18-1 


0-70 







1,744 


15-2 


0-59 







1,917 


12-4 


0-48 






' 1,320 


20-1 


0-42 







1,660 


18-3 


0-38 


__ 




1,820 
1,990 


16-8 


0-35 






131 


0-27 


1 



Repetitions of Stress." L. B. Turner, Engineering, 11th and 25th Aug. 

. II *' On the Resistance of Iron and Steel to Reversals of Direct Stress." 
Stanton and Bairstow. Proc. Inst. Civil Engineers, 1906, vol. clxvi. 

1[ " On a Throw Testing Machine for Reversals of Mean Stress." Professor 
Osborne Reynolds and J. H. Smith. Phil. Trans. Royal Society, 1902. 




96 THE STRENGTH OF ^L\TERIALS 

standpoint this seems to be a somewhat too mmidane de- 
scription of the engineer's vocation, we must not forget that 
the most scientific construction is the one which best fulfils 
the conditions for the least cost. 

There is in reality no conflict between theor}^ and practice 
in designing ; each has its own place, and each is dependent 
on the other. The theory will tell us what is the best design 
as far as the economical arrangement of material goes. The 
best-designed structure is one which would be about to 
collapse at all sections at the same time ; or, in other words, 
the various parts are so designed that the stresses in them 
are equal. This is all that the theory sets out to do. Practice, 
on the other hand, determines whether the theoretical design 
is in reality the cheapest in the end. Questions of workman- 
ship, cost of erection and upkeep have to be considered, and 
it is only by balancing these with the theory that the really 
scientific design is obtained. 

In dealing with the theoretical side of design we must 
never forget that, if we are to be guided by theory at all, we 
should see that we use the best theory. The disdain for 
theory that ultra-practical men often possess is largety due 
to the fact that their theoretical knowledge is not sufficiently 
comprehensive; they have not realised the conditions which 
have to be fulfilled before a certain theory is applicable, and 
so they probably use some formula for a case for which it was 
never intended. 

Another point to be remembered is that practical rules for 
use in design are not necessarily sound because the machines 
or structures resulting therefrom satisfactorily fulfil their 
function. Such rules may make the design much heavier, 
and therefore much more costly, than necessary. Our aim 
in the theoretical investigations should be to eliminate as 
many uncertainties as possible, and not to be merely content 
with erecting something which will stand. 

Commercial Aspect of Design. — If the word "scientific" 
is used in its best sense, the commercial aspect differs very 



REPETITION OF STRESSES 97 

slightly from the scientific as23ect. There are certain points, 
however, that we would like to deal with which point to 
the necessity of considering the merely commercial aspects. 
First, there is the question of the sizes of sections adopted. 
Care should be taken that as much as possible is used of the 
same section, and that such section should be easily obtain- 
able. The cost of a given structure may be increased largely 
because a section is specified which has to be rolled specially 
— although sections figure in makers' catalogues they are 
not always readily obtainable. In riveted work, too, much 
additional cost is often involved by an unnecessarily irregular 
pitch of the rivets, and fancy forms of cleated connections are 
often shown which have no advantage over the simple forms. 

The designer should avoid curved lines in structural steel- 
work wherever possible in his design. It costs a lot to cut 
plates to a curve, and there is generally no reason for them. 
Some might urge that curved forms are more pleasing to the 
eye, and some go as far as to put cast-iron rosettes on the 
plates of plate-girders. But it is better to agree that no steel 
structure is artistically beautiful, and that to attempt to 
decorate it by curved gusset plates and rosettes is to make it 
really more ugly, because it has cost more and is still an eye- 
sore to the artist. There is, also, a theoretical objection to 
curved members, viz. that the loading on such bars is eccentric, 
and stresses are therefore much increased. 

Where practice necessitates our putting theory aside some- 
what, we should always keep this in mind in our calculations. 
For instance, theoretically the centre-line of the rivets in 
a T section should coincide . with the centroid line of the 
section. In practice this is impossible, as the head of the 
rivet could not then be closed. But we must remember in 
designing that the load is eccentric and that due allowance 
must be made for this. 

We shall in this book be concerned only with the strength 
of structural details and machine parts, but it may be pointed 
out that in designing castings the pattern-maker must be 



98 THE STRENGTH OF MATERIALS 

considered and that ease in machining must be ke2:)t in mind ; in 
fact, everything must be done Avhich will save needless expense. 

Working Stresses and Factor of Safety. — The question 
of the working stresses to adopt in practice is of the utmost 
importance, and if our design is to be of any real value we 
must have clear ideas as to such working stresses. 

In dealing with working stresses we often speak of the 
factor of safety. This ma}' be defined as the factor b}' 
which the working stresses may be multiplied to give stresses 
which will result in failure. This phrase is one which is often 
used glibly without any real meaning ; and it has been sug- 
gested that in many cases it would be better called the factor 
of ignorance. If we design a structure with a factor of safety 
of four, say, we certainly do not as a rule mean that the 
structure could bear four times the load without failure. This 
is because there are certain contingencies that we do not 
allow for in our design. Our aim should be, however, to 
make our calculations so that the factor of safety has as exact 
a meaning as possible. This can be done only by choosing 
our working stresses skilfully and by making allowance for as 
many points as possible. For steel-work it is common to 
adopt as a working stress in tension one-quarter of the 
breaking stress in tension and to say therefore that the factor 
of safety is 4. Man}' designers forget, however, to make the 
due allowance for live or variable loads. The basing of the 
factor of safety on the breaking stress is also open to a very 
serious objection, viz. that the elastic limit of the material 
is the point which really determines the safety of the struc- 
ture. If the stresses are above the elastic limit, failure is 
almost certain to ensue, especiall}' in the case of compression 
members or struts. It would, therefore, be better to base the 
working stresses on the elastic limit or the yield point, since 
in pure tension the two points are close together, and the 3'ield 
point is much easier to measure and specify for a definite 
minimum value of such limit in the steel. The point com- 
monl}^ urged against this method of procedure, viz. that the 



REPETITION OF STRESSES 



99 



elastic limit is a much more variable quantity than the 
breaking stress — seems to us to be one in favour of its adoption. 
It is certain that stresses beyond the elastic limit are very 
dangerous, and if this quantity is a variable one we ought to 
know it for the material that we are using, and base our 
working stresses on it accordingly. We would suggest that 
the dead-load or static working stress should be taken as 
one-half of the natural elastic limit. 

The following tables of stresses may be used for obtaining 
the usual working stresses adopted for dead loads in design : — 







Working Stress. 




Material. 








Dimensions of 

Stresses. 


Tension. 


Compression. 

r 7 
(bending) 

1 ^ 

'^ (direct) 

J (bending) 

4 

^ (direct) 
4 


Shear. 


Mild steel * . . . 


7 


5 


tons per sq. in. 










Wrought iron . 


5 


4 


>J J5 


Cast iron . 


1 

2 


i 


5J J> 


Oak . . . . 
Pine, yellow 


16 

3 


13 

6 

r 600 
J (bending) 

500 
'^ (direct) 
35 


5 

(across grain) 

3 
(across grain) 


cwt. per sq. in. 

5) ;? 


Cement concrete 1 
1:2:4 . . j 


60 


60 


lbs. per sq. in. 


Granite .... 


— 


— 


tons per sq. ft. 


Sandstone . . ) 
Yorkstone . . f 


— 


20 


— 


JJ J5 


Limestone . 


— 


15 


— 


; J 5j 


Brickwork in 










cement mortar 


'5 


8 


- 





(adhesion) 
in lime mortar , '4 

(adliesion) 



* Many authorities allow J ton more for each of the stresses in mild 
steel. We have already seen that according to one theory the working 
shear stress for ductile metals should be half the tensile strength, viz. 
3'75 tons per sq. in. for mild steel. This figure is not, however, in 
common use. 



100 THE STRENGTH OF MATERIALS 

Allowance for "Live" Loads or Variable Loads. — 

There are two priiicijial methods of allowing for live loads 
which are in effect the same. 

(a) Equivalent Dead-load Method. — According to this 
method the static stresses are used and the loads are in- 
creased to give the equivalent dead load. The ways for 
allowing this, are — 

(1) equivalent dead load = dead + 2 live load. 
This may be called the dynamic formula. 

(2) equivalent dead load 



= w. 



n r + yJ n^ r'^ + 4: (w — -^J 



2 

where r is the variation of load, and w is maximum load, 
n being a constant which may be taken as I'S for steel. This 
formula is deduced from Unwin's formula for Wohler's 
experiments. 
For steel we get 



1-5 r + J 2-25 r^ ^ 4.(w - 



2) 

w,. = 



2 
When the variation is from zero to a maximum, we have 

r = w. 

Then w, = 21 w. 

(3) equivalent dead load = maximum load + variation. 

(6) Variable Working Stress Method. — According to 
this method the working stress is varied according to the 
relative amounts of live and dead loads. 

The common ways of allowing for this are — 

(1) Launhardt-Weyrauch method. 

^^^ - . / /' minimum load \ 

Worknig stress = —^ 1 + o " • 1 ^ 

° 1*5\ 2 X maximum load/ 

/ being the static or dead-load working stress. 

(2) Dynamic method. 

Working stress = rV — . — ^ , / being as before. 

total load 



KEPETITION OF STRESSES 101 

Take as a simple numerical example the case of a member 
of a roof truss in which the dead load is a tension of 5 tons, 
and the wind on one side causes a tension of 2 tons and on 
the other side a compression of 1 ton. The various methods 
give the following results : — 
(a) (1) Equivalent dead load = 5 + 2 x2 = 9 tons. 

3\2 
2 



1-5 X 3 + J 2-25 X 9 +4(^7 





y^) 


5> 


5J 


. 


8-2 tons. 


2 




(3) 


J5 


33 


= 


7 + 3 = 


10 tons. 


(b) 


(1) 


Working 


stress 


= 


6/ 

7 






(2) 


5J 


33 


= 


/ 


7/ 
9 



Assuming the material to be mild steel. 
[h) (1) gives working stress = 6 tons per square inch. 
(2) ,5 5, = 5*4 ,, J, ,, 

Taking the material as mild steel, the requisite number of 
square inches in the sectional area of the tie are — 

9 
(a) (1) = 1-28 square inch. 



(2) 7 


= 117 


(3) '^^ 


= 1-43 


ib) (1) I 


= 117 


^^' 5-4 


= 1-30 



If consideration of variation of stresses be neglected alto- 
gether, we should have — area = =1 square inch. 



CHAPTER IV 

RIVETED JOINTS ; THIN PIPES 

Forms of Rivet Heads. — The most common forms 
of rivet heads and their usual proportions are shown in 
Figs. 44, 45. 

For structural work the snap-headed rivets are most usual, 
but countersunk rivets are used where necessary to prevent 



CUP or SNAP HEAO 




CONiCAL HEAD 




PAN HEAD 



COUNTE R SU'SK HE A O 



Figs. 44, 45. — Forms of Rivet Heads. 



projections from the surface of the plate. Snap-heads take a 

length of rivet equal to about IJ times the diameter. 

It is usual in practice to adopt a diameter of rivet when 

cold equal to one-sixteenth of an inch less than the diameter 

of the hole, but in all calculations the diameter of the rivet is 

taken as being equal to that of the hole. 

Forms of Joints. — (a) Lap Joints and Butt Joints. — 

102 



RIVETED JOINTS; THIN PIPES 



103 



In the lap joint the plates overlap as shown in Fig. 46. This 
form of joint has the disadvantage that the line of pull is 
such as to cause bending stresses, tending to distort the joint 
as shown. 

In the hutt joint the edges of the plate come flush, and 
cover plates are placed on each side as shown, the thickness 
of each cover plate being usually five-eighths that of the main 
plates. In this form of joint the pull is central, so that there 
are no bending stresses. 

In the single cover joint, which is a cross between the lap 



I 



' I 



^^ 



I 
T 




4- 



% 



I 




LAP JOINT. 



BUTT JOINT. 

Thickness of Cover 



SINGLE COVER JOINT. 

5 t 

Thickness of Cover — - 

4 



Fifi. 40. — Forms of Riveted Joints. 



joint and the butt joint, there are bending stresses developed, 
tending to distort the joint as shown. 

It IS clear from the above that the butt joint should be 
adopted wherever possible. 

(6) Chain Riveting and Zig-zag or Staggered Rivet- 
ing. — The different rows of rivets in a joint may be arranged 
in chain form or zig-zag form, as shown in Figs. 47, 48. As 
we shall see later, the zig-zag form is more economical, and 
should be used whenever possible. 

The essential feature of zig-zag riveting is that the rivets 
in alternate rows are displaced laterally by half the distance 
between the rivets, i. e. by half the pitch of the rivets. In 



104 



THE STRENGTH OF ]\UTERIALS 



the form shown the joint is in a tie bar of a bridge and the 
rivets form a triangle ; it is common in boiler and like rivet- 
ing to make the pitch of the outermost of three rows twice 
that of the innermost row; the result is a special kind of 
zig-zag riveting which Ave may call triangular riveting. 

Methods by which a Riveted Joint may Fail. — A 
riveted joint may fail in any of the following ways : — 









o 


o 


o 


o 


o 


o 


o 


o 


o 





o 


o 











Fig. 47. — Chain Rivetine:. 



Fic 4S. — Zig-zag Riveting. 



(1) By tearing of the plate. 

(2) By shearing of the rivets. 

(3) By crushing of the rivets-. 

(4) B}' bursting through the edge of the plate. 

(5) By shearing of the plate. 

Fig. 49 shows these methods of failure. 

(4) and (5) are allowed for by the following rule : The 
minimum distance between the centre of a rivet and the 
edge of the plate is 1 J (h where d is the diameter of the rivet. 

If this rule is adhered to the joint will always fail first in 
one of the ways (1), (2), (3). 

The aim in designing a joint should be to make the force 
necessary to cause failure in the various ways equal. 

We will now consider the various ways of failure in detail, 
taking in each case a strip of plate equal to the pitch of the 
rivets. 

(1) Tearing of the Plate, — In this case the width along 



RIVETED JOINTS; THIN PIPES 



105 



which fracture will occur is {p — d), and as the thickness of 
the plate is t, the area of fracture ^ {p — d)t. 




VI 



/. 



/ 



S 






P 




/ 

z 



^ 



'< ?■ 



^^ 




DOUBLE Sfi£/^fi , 



P 




-A 



,p 

5 





Fig. 49. 



♦ P 



(2v- 



-A 




Therefore, if /, is the safe tensile stress in the material, the 
safe load which the joint can carry is equal to 

V =tt{V-d)t (1) 



U)() THE 8TRENGTH OF MATERIALS 

(2) Shearing of the Rivets. 

Til the case of single shear, the area sheared ^ 

double ,, 



4 

2jr_a2 
- 4 - 



Therefore if /, is the safe shear stress on the rivet, the safe 
forces on the joint as regards shear are respectively 

P = * 4 J 

(3) Crushing or Bearing of Rivets. — In this case the 
crushing or bearing area is taken as the diameter of rivet 
multiplied by the thickness of the plate, i.e.d x t. There- 
fore, if /„ is the safe bearing stress on the rivet, the safe force 
on the joint as regards bearing is equal to 

V^j^^.d.t (3) 

The values of j, and s may be taken as given in Chap. III. 

For /,„ 10 tons per square inch may be taken for mild steel, 
and 8 tons per square inch for wrought iron. These figures are 
higher than for ordinary compression, and are obtained from 
the results of experiments. 

For structural work the strength of the joint as regards 
bearing will often be less than as regards shear, because the 
plates are often thin compared with the diameter of the 
rivet. 

Efficiency of Joint. — The efficiency of a joint is the 
percentage ratio of the least strength of a joint to that of a 

solid joint, i. e. 

-p,^ . _ _ Least strength of joint 

•^ ~ ^ "" Strength of solid plate 

Diameter of Rivets. — For the most economical joint the 

* A Board of Trade rule states that this should be taken as ^ , 

4 

and this rule is often though not universally adopted. This figure is 

based upon the results of tests. 



RIVETED JOINTS; THIN PIPES 



107 



diameter of the rivet should be such that the shear and 
bearing strength are equal. 

Unwin's Formula, which is in common use, gives 

d = l'2Vt (A) 

For single shear we have 

shear strength = a -^^ 



bearing strength =^ dt .f^. 
If these are equal 

c^ = '^^Z" = 2-54^ (if f,^2s) 



Thickness of Plate (inches). 

Fig. 50. — Diameter of Rivets. 



(D) 

























y' 




































































y 






















y 


y 






















y 






















^ 






















y 










^^-^ 












y 


^ 






^ 


^-^ 


^,^ 










y 


y 






^^-^ 


^ ^ 


^^ 


















,,.^-^ 


^..^ 


^ . 












/ 




^ 


:^^ 


■ — ^ 














y^ 


^^ 


^:^ 


^^^^-^ 
















^ 


^^-^ 


<^ 


















.^ 


-^p^ 






















^ 


^ 

























A 


P 


I 




"3 








1 


1 

4- 


li 



4 



Z 






Ph 



Oq 



For double shear we shall have 

2 7r6Z2 

^ -s = dt /,. 



d = 



TT S 



or on the Board of Trade rule 

dtf,, 



1-75 TT 6^2 



\21t 



32 t 



TT . 1-75 5 7 TT 



(B) 



(C) 



108 THE STRENGTH OF MATERIALS 

These values are plotted in Fig. 50. 

It is clear from this figure that for large thicknesses of plate 
for single shear the theoretical value gives diameters of rivet 
which are impossible in practice. 

Numerical Examples. — (1) A tie bar in a bridge consists 
of a flat bar of steel 9 inches wide by l^ inches thick. It is to 
be spliced by a double butt joint. Determine the diameter of 
the rivets and their number, and give sketches showing the proper 
pitch and arrangement of the rivets. {B.Sc. Lond.) 

According to Unwin's formula d = 1'2VT= 1'34 inches. 
This is, however, rather high for practice, and so we will 
adopt d =^ \ inch. 

Assuming that the rivets are arranged in zig-zag fashion, 
the strength of the joints against tearing through the outside 
rivet is equal to 7 (9 — 1) . IJ = 70 tons. 

Shear strength of each rivet = 5 . .^ • (1)^ = 7*85 tons. 

. • . Number of rivets required for shear 

= ^|g = 8-93 = say 9. 

Bearing strength of each rivet = 10 x 1 x IJ = 12*5 tons. 

70 
.-. Number of rivets required for bearing = ^^.r ^ ^^y ^• 

9 rivets would thus be ample as regards bearing. 

The joint would then be arranged as shown in Fig. 51, the 
centre two rows being chain-riveted. 

We will now consider the strength of this joint under various 
ways of failure. 

If the plate tears along the line A a, the force necessary to 
reach the safe limit of stress is, as we have shown above, 
70 tons. 

Now suppose that the plate tore along B b, shearing off the 

rivet in A A. 

5 
Then strength of line bb = 7(9-2)= 61'25 tons 

Strength of one rivet = 7*85 tons. 



RIVETED JOINTS; THIN PIPES 



109 



. • . Total strength against failure along b b 

= 61-25 + 7-85 = 69-1 tons. 

Now suppose plate tore along c c, shearing off the three 
rivets. 

5 

Then strength of line cc = 7(9 — 3). = 52-5 tons. 

Strength of three rivets = 23' 55, 

. • . Total strength against failure along c c 

= 52-5 + 23-55 = 76-05 tons. 

D C B /^ 

Rit/eis I Diam 




-1 1 — u 



J — u 



tT^ 



J L 



i"rf;d 



Cover P/oTo. L- 



3 

Fig. 51. 



Finally, suppose cover plates tore along d d, then strength 

7 



= 7(9 - 3). 2. 



73-5 tons. 



From the above we see that the weakest section is along 

B B. 

mi nn ' p • • ^ Least strength of loint 

Inen emciency 01 lomt = c^, .i — r.^ — ,• v \ 

Ibtrength of sohd plate 

691 69-1 



9 X li X 7 



78-8 



87'8%. 



llu THE STRENGTH OF MATERIALS 

If instead of zig-zag riveting we had adopted chain riveting 

with three rows of three rivets (9 in aU) the least strength 

would be (9 - 3) IJ x 7 = o'2o tons. 

52 ' 5 
.'. efficiencv of joint = _ , = 667 ^o- 

If we had four rows of chain riveting with two rivets in 
each row (8 in all), the least strength would be (9 — 2)1 J 
X 7 = 6T2o tons. 

.•. efficiencv of joint = ^^^ = 77*7 ^o- 

too 

The above shows that the zig-zag riveting is considerably 
more efficient than the chain riveting, and is therefore more 
economical. 

(2) Design a douhle-rivefed lap joint to connect two steel 
plates J in. thick unth steel rivets, the tensile strength of the 
plates before drilling being 30 tons per sq. in.; the shearing 
strength of the rivets 24 tons per sq. in. : and the compressive 
strength of the steel 43 tons per sq. in. Find the efficiency of the 
joint. {A.M.I.C.E.) 

For J in. plates Unwin's formula would give 
d = 1"2 \ "5 = 'So in., say | in. 

The joint is a double-riveted lap, therefore there will be two 
rivets in single shear in a width of plate equal to the pitch. 

. • . Strength against tearing per pitch 

= /• {p-d)t 

= 30(p-(/).^ = \o{p-d)..{\) 

. ' . Strength against shearing per pitch 

_ 2 - r/2 
~ ' ■ 4 
24 .2- 



28-9 tons. 



II these are equal 15 ( p — „ ) = 289 



28-9 _ 7 
15 8 
- 1-93 + -87 - 2-80,sav3ins. 



•■• ^^= 15 8 



to 



RIVETED JOINTS; THIN PIPES 111 

The bearing stress for a force of 28*9 tons would be equal 

28-9 



■q" X 9: X ^ 



33 tons per sq. in. 



7 1 
the bearing area of each rivet being q x -^ = *437 sq. in. . 

This is less than the allowable value of 43 tons per sq. in., 
showing that a larger diameter of rivet might be used with 
greater economy, but | in. diameter is in most cases more 
suitable in practice. 

The efficiency of joint in this case is equal to 

28-9 _ 28-9 _ 
30 X 3 X i ~ "45~ ~ ^* ^ /^ 

The joint then comes as shown in Fig. 52. 






^ 



t) 



f 



:i 



Fig. 52. 

(3) A steel-plate tie bar in a bridge is subject to a tension due 
to dead load only 0/ 16 tons. The stress due to live load only 
varies from 36 tons tension to 10 tons compression. The tie 
bar is | in. thick and is to be joined to the side plate of a girder 
by means of a % in. gusset plate and double-cover butt joint- 
Select suitable working stresses and design the joint, arranging 
the rivets so that the tie bar is weakened by only one rivet section. 
{B.Sc. Lond.) 

The maximum load in this case is 36 + 16 = 52 tons, and 
the minimum load 16 — 10 ?= 6 tons. 



112 THE STRENGTH OF MATERIALS 

Using the Launhardt-Weyraiich formula, we have 

„. , . ^, f /i , i^ii^- stress \ 

Working Stress = v r 1 + « 

° I'o \ 2 max. stress/ 

= l4(l + 104) = -™^/ 

This gives a tensile stress of 4*93, say 5 tons per sq. in. ; 
a shear stress of 3"52^ say So tons per sq. in. ; and a bearing 
stress of 7 tons per sq. in. 

According to UnAvin's formula d = 1'2 V '15 = r04 in., 
but for practical reasons | in. would usually be adoi)ted. 

We now require to find the necessary width of the tie bar. 
Let this be w. 

Then Iw — ^ ) t ^^ ^^^ equivalent cross-sectional area. 

/ 7\ 3 

.' .(w — -q) • A- 5 must be equal to the maximum pull 

of 52 tons. 

V 8/ 3x0 

.- . w =^ 13*89 + '875 = say 15 inches. 
The strength of each rivet in double shear is equal to 

2 7r nv 



, . , ^ , . 3-5 = 4-22 tons. 
4 \8/ 

52 

.' . Number of rivets required for shear = ,^^ = 12*3. 

We will use 14, as they give the best arrangement. 
The strength of each rivet in bearing is equal to 

3 7 

-7.^-7^ 4-58 tons. 

4 8 

.• . 14 rivets will be ample for bearing. 
The joint is then arranged as shown in Fig. 53. It is very 
important in such joints that the centre line of the rivets 
should coincide Avith the centre line of the tie bar, or else the 
pull in the bar would be eccentric. In such joints, therefore, 
the rivets should always be arranged symmetrically with 
regard to the centre line of the tie bar. 



RIVETED JOINTS; THIN PIPES 



113 



(4) Find the number of rivets necessary to the gusset plates, 
etc., at the base of a steel stanchion to the stanchion proper, the 
load carried being 150 tons. The diameter of the rivets is | in. 
and the thickness of the plate J i'^- 

The rivets usually have to be designed in such cases so that 
they will carry the whole load, so that if the stanchion itself 




Fig. 53. 

does not bear on the base plate the rivets will distribute the 
load satisfactorily. 

The strength of each _ tt 

rivet in single shear 4 

The strength of each _ 7 

rivet in bearing 8 



v8 



3-01 tons. 



10 = 4-37 tons. 



150 



.'. Number of rivets necessary = „ , = 50 nearly. 

Some Practical Considerations in Riveted Joints. 
— Punching and Drilling of Rivet Holes. — It is quite 
common in this country for specifications to state that rivet 



114 THE STRENGTH OF MATERIALS 

holes must be drilled out of the solid. Punching is known to 
injure to some extent the material in the neighbourhood of 
the hole, and is thus often objected to. The extent to which 
punched holes weaken a structure such as a plate girder 
compared with drilled holes does not appear to have been 
satisfactorily determined, although such determination from 
a practical point of view^ would seem to be absolutely neces- 
sary, since there is an increase in cost entailed in drilling the 
holes. In recent years punching machines and means for 
obtaining an accurate pitch of the holes have been improved 
considerably, and when we consider the increased cost of the 
drilling, punching is preferable in many cases. In recent 
years " gang " or multiple drilling machines have been 
introduced which lessen the cost of drilling; one great ad- 
vantage that drilling possesses is that the plates to be joined 
can be clamped together and drilled right through, thus 
ensuring accurate registering of the holes. A good com- 
promise is to punch the hole i to J inch less than required, 
and to reamer out to size, the damaged metal being thus 
removed ; but this is considerably more expensive than plain 
j)unching. A method of allowing for the damage of metal 
due to punching which has been suggested, and which we 
consider preferable, is to add J inch to the diameter of the 
hole in calculating the tearing or tensile strength. This adds 
very little to the size of the plate and saves in cost of pro- 
duction. The point that should be very carefully seen to is 
that the holes are accurately pitched, so that the holes will 
register well when the parts are assembled, and will not 
require excessive drifting as is the case when the spacing of 
the holes is inaccurate. It is probable that many more 
joints are unsatisfactory because the rivets do not fill the 
holes, owing to the latter not registering accurate^, than 
because the metal has been injured owing to punching the holes. 
There is considerable friction between the plates in a 
riveted joint, but this is not allowed for in calculations of 
the strength. 



RIVETED JOINTS; THIN PIPES 



115 



Pitch and Spacing of Rivets. — In order to prevent 
moisture getting between the plates and causing bulging due 
to rusting, or to prevent local buckling in the case of com- 
pression members, it is common to stipulate that the pitch 
of rivets shall not be greater than 6 ins., or sixteen times the 
thickness of the thinnest plate. The designer should re- 
member that pitches from 3 ins. upwards, increasing by half- 
inches, should be used, and odd fractional pitches avoided, 
except where absolutely necessary. As far as economically 
possible, the same pitch should be used throughout, and in 
many cases, for girder work, etc., 4 ins. is used unless special 
conditions require a different pitch. 



Working Strength of Steel Rivets 



Diam. 


Area in 
sq. ins. 


Strength 
in single 
shear at 
5 tons 
per sq. in. 




Bearing Strength at 10 tons per sq. 


in. 


of 
Rivets 


Thickness in ins. of plate. 


in ins. 




tV 


3 

8 


7 


4 


9 
To 


1 


1 1 


'^ 


f 


•1104 


•55 


M7 


1-41 


1^64 


1^87 


211 234 


2^59 


2^81 


* 


•1963 


•98 


1^56 


1-87 


2^18 


2^50 


2-81 3^12 


343 


375 




•3068 


153 


1^95 


2^34 


2^72 


3^12 


3-51 1 3^90 


4^30 


4^68 


3 

•i 


•4418 


2-21 


2^34 


2^81 


3^27 


3^75 


4-21 4^69 


5^16 


5^63 


g 


•6013 


301 


2-72 


3^27 


3^82 


4-37 


4^91 5^46 


6^02 


6^56 


1 


•7854 


3^93 


312 


375 


4^37 


5^00 


5-62 6^25 


6-87 


7^50 



Thin Pipes and Cylinders. — Suppose that a thin cylinder 
of diameter d, Fig. 54, and thickness t, is subjected to a pressure 
of intensity p. This pressure will tend to burst the pipe along 
a longitudinal section, and the pressure on the ends will tend 
to cause failure across the circumferential section x x. In 
thick pipes the stress will vary across the section and is dealt 
with in Chap. XVII. 

Longitudinal Section. — Consider a length I of the pipe. 
Then the radial pressure p at any point can be resolved into 
a component o b parallel to any diameter under consideration 
and a component a b normal to it. The resultant normal 
force will be the same as the pressure acting over the 
diameter. 



116 



THE STRENGTH OF IVIATERIALS 



If /, is the tensile stress across the section, we have 

. • . Force tending to burst pipe = p x area = p d I. 
Force resisting bursting = /, x area. 

= /, X 21 1 {It on each side). 
, p d I 

}) d 

^ 2t 

This stress ft is often called the hoop stress. 



/^^ 



'>!}>»>;".:/•'! irrr. 



ii 



v..',;. .',"/A7///,/»/ ,'.'- '-ns 



^ 



Tzz: 



t 



l^ / — >i 



-^^ 



(1) 





Fig. 54. — Stresses in Thin Pipes. 



Circumferential Section. — If // is the tensile stress 
across the section x x, we have 

Force tending to cause failure = p x area of end. 

Force resisting failure = // X area 

= // X TT d t 

(because the pipe is thin). 



RIVETED JOINTS; THIN PIPES 117 

• • A =P -4- - Trdt 

= 1^ (2) 



1 



/< 



Therefore the stress across a longitudinal section is twice 
that across a circumferential section; for this reason longi- 
tudinal joints of boilers are made stronger than circum- 
ferential ones. 

* Equivalent Stresses on Strain Theory. — On any small 
cube of the material with sides parallel and normal to x x, there 
will be a hoop stress /„ a longitudinal stress \ ft, and a radial 
stress which is jj on the inside and o on the outside of the 
tube and may generally be neglected. 

. ' . Strain in longitudinal direction = -U — 4V, 

E V 2 



7/, f 1 



7 
. • . Equivalent hoop stress = ^ ft 

o 

Similarly stress on circumferential section = '^ -\- q f^ 

4 

On the equivalent strain theory, therefore, the pressure on 
the ends strengthens the pipe. 

Numerical Exaiviple. — A holler 1 ft. Q ins. in diameter has 
to sustain a pressure of 80 lbs. per sq. in. If the efficiency 
of the joints is 70 per cent, and the safe stress is 4 tons per sq. 
in., find the thickness that the boiler should have and the necessary 
pitch of rivets on the longitudinal butt joint. 

. _pd ^ 80 X 90_ _ 90 

2 / ~ 2 X 4 x" 2240 ~ 224 
Efficiency of joint = 70 per cent. 



118 THE STRENGTH OF MATERIALS 

rru- 1 ^ 1. ^ X 100 90 X 100 

.'. Imckness must be — ^^ — — c^^. ^r. = "Svo in. 

70 224 X 70 

say f in. 

Diameter of rivets = 12 V t = 1 in. nearly. 

Efficiency = 70 per cent. 

T , . p — d „ 
.' . Intension — = "7 

P 
i.e. p — I = '1 p 

.'. Sp = 1 

p = 3" 3, say 3 J ins. 

TT 

Strength of each rivet in shear = . x l'15d^ x 5 

= 6'9 tons. 

5 

Strength of plate per pitch = 7 x 2*25 x 

o 

= 9-85 tons. 

9*85 
.• . Number of rivets required per j)itch = /,.q =2 (as whole 

numbers only are possible). 

. • . A double row of rivets are required on each side of the 

joint with a pitch of 3 J ins. 

Collapse of Thin Pipes under External Pressure. — 

If thin pipes are subjected 
to external pressure there 
will be hoop compression 
stresses which may be cal- 
culated by the same formulae 
as we have obtained for the 
hoop tension due to internal 
pressure. If there is any 
inequality in the pipe bend- 
ing stresses will be induced 
which will cause failure by 

collapse before the crushing strength of the material is 

reached, this collapse being similar to the failure of columns 

by buckling. 

Fig. 55 shows some of the forms of collapse of such tubes. 






RIVETED JOINTS; THIN PIPES 119 

Fairbairn's Formula. — The first well-known experiments 

on the subject were made about 1860 by Fairbairn. His 

formula is 

806,300^219 

p = collapsing pressure in lbs. per sq. in. 
d = diameter of tube in inches. 
t = thickness ,, ,, 

L = length of tube in feet. 

Board of Trade Rule. — 

Safe pressure = j . , for single-riveted lap-welded tubes. 

= for welded or double-riveted butt- 

(L + l)a jointed tubes. 

Stewart's and Illinois Experiments.* — These recent 
experiments were made with great care and proved that 
except for tubes of length less than about 5 diameters the 
collapsing pressure is practically independent of the length, 
so that Fairbairn's and the Board of Trade Rules are not 
applicable. 

Fig. 56 shows the results of Professor Stewart's experiments 
for 4 and 7 in. tubes ; it is clear from this that beyond a 
certain thickness the collapsing pressure is practically pro- 
portional to the thickness. The Illinois experiments confirmed 
this. 

The following formulae are given : — 

Very thin tubes ( 7 <C "03 j — 



Brass: ^ = 25,150,000 (-0 (Illinois) 



Cold-drawn seamless steel : p = 50,200,000 ( -, ) (Illinois) 

Stewart finds that the same formula holds for lap-welded 

Bessemer steel tubes. 

* Am. Soc. Mech. E. 1906 (Reid T. Stewart) ; Illinois University 
Bulletin, 1906 (A. P. Carman and M. L. Carr). 



120 



THE STRENGTH OF MATERIALS 



Tubes in which , "> 03 — 
d 



Brass : p = 93365 . - 2474 (Illinois) 



d 



4" 



cooo 


































/ 








fiSOO 
































.<(;/. 








5C0O 


































'-t^ 








SlOO 








































6200 
6000 
1800 
4C00 
«iOO 
































/ 




































47 






































■^'k 


/ 














































































1 
















J3 
























/ 
















4000 


c- 
























/ 
















3600 
























/ 


















Q. 






















/ 














. 


^ 


34 00 
3200 
3000 
2800 
2600 


-2 
►J 




















I 


















?' 




















/' 


































I 


r 












■< 


^ 






L 


















1 












-f^o 


\y 








3 
















J 


V 










+ 


/ 










2400 
2200 
2000 
1800 


a 
















A 












y 


-^ 










'7; 

a. 














k 


A. 










^ 


/ 












o 














t 
1 












/^ 
























i 


/ 










Y 
















1400 














/ 










/ 






























\l 








y 


^ 


















1200 




















/ 


/ 




















800 












/ + 








/ 






























/ 


J 




H 


/ ■¥ 
























600 
400 
















/ 


^ 






























* 




,^ 


y 
































^,.- 


r- 


.-' 


'' 




Thick 

1 


nes£ 


of Wall', in 

1 1 


Dec 


niais of 

1 1 


an lnc^ 

1 


1 




1 




.(J 


2 .C 


)4 .0 


6 .i 


3 .] 


.1 


2 .1 


i .1 


.1 


8 A 


.2 


2 .2 


4 .i 


C .S 


8 .S 


.3 


2 .J 


14 .2 


c .s 


18 .40 



Fig. 56. — Collapse of Tubes. Stewart's Experiments. 
Seamless cold-drawn steel: p = 95520^^ - 2090 (Illinois) 

Lap-welded Bessemer steel: f = 83270 ^ - 1025 (Illinois) j 

1386 (Stewart) 



-f 86670 ^ 



d 



CHAPTER V 

BENDING MOMENTS AND SHEARING FORCES 

ON BEAMS 

Definitions. — The shearing force at any point along the 
span of a beam is the algebraic sum of all the perpendicular 
components of the forces acting on the portion of the beam 
to the right or to the left of that point. 

The bending moment at any point along the span of a beam 
is the algebraic sum of the moments about that point of all 
the forces acting on the portion of the beam to the right or to 
the left of that point. 

As the beam is in equilibrium under the forces acting on it, 
the algebraic sum of the forces at any point, and of the 
moments of the forces about the point, acting on both sides 
must be nothing; so that we shall get the same numerical 
values for the shearing force and bending moment from 
whichever side we consider them, but they will be opposite in 
sign. We will, wherever convenient, consider the shearing 
force and bending moment of the forces to the right of the 
section, and we will take an upward shearing force and an 
anti-clockwise bending moment as positive, the downward and 
clockwise being taken as negative. 

Bending Moment and Shearing Force Diagrams. — 

If the bending moment and shearing force at every point of 

the span be plotted against the span and the points thus 

obtained be joined up, we shall get two diagrams called the 

Bending Moment (B.M.) and Shear diagrams, and from these 

diagrams the values of these quantities can be read off at 

121 



122 THE STRENGTH OF MATERIALS 

any point of the span. We will consider the forms of these 
diagrams for various kinds of loading and for various ways of 
supporting the beam, but will only consider beams with fixed 
loads. We will use Mp and S,. to represent respectively the 
bending moment and shearing force at a point p. 

B.M. AND SHEAR DIAGRAMS WITH FIXED LOADS 

A. Cantilevers,* i.e. beams fixed at one end and free at 
the other, the loads being all at right angles to the length 
of the beam. 

Case 1. Cantilever with One Isolated Load. — Let a 
cantilever, fixed at the end b, Fig. 57, carry an isolated load 
W at the point a, at distance I from b. Consider any point p 
at distance x from a. 

Then we have S,. = W. 

This is constant throughout the span. 

.*-. Shear diagram is a rectangle of height W. 
Again M, = W x x 

This is proportional to x. 

.' . B.M. diagram is a triangle whose maximum ordinate is 
W I, this being the bending moment at the point b. 

Case 2. Cantilever with Two Isolated Loads. — Since 
the B.M. and shear at any point are defined as the sum of 
the moments and the forces to the left of that point, it 
follows that the B.M. and shear diagrams for a number of 
loads can be obtained by adding together the diagrams for 
the separate loads. In the present case, in which we have 
loads Wi and Wg at distances l^ and /g from the fixed end, 
the diagrams are obtained by adding together the separate 
diagrams as shown in Fig. 57 (2). 

Case 3. Cantilever with Uniform Load. — Let a uni- 
formly distributed load of w tons per foot run be carried by a 

* According to our convention the shears and B.Ms, for all the cases 
of cantilevers that we consider are negative. There is, however, no 
need to give the sign, unless both positive and negative values occur in 
the same beam. 



BENDING MOMENTS AND SHEARING FORCES 123 



cantilever a b of span I. Consider a point p at distance x 
from the free end a. Then 



?<r- 



t 
WL 



y/}i^//'^//. 




Rending Moment 

/so/ated Load 



v 




^1 



w, \ 

V 



^Z 



z, 



y//^!^^/M/^/y/777P7A 




© 

TTvi? isolated Loads 



toOQCQCCpD 




' fjarobola { 'X 

Un ijo rm L ooid 

Uniform and isolated Loads 

Fig. 57. — B.M. and Shear Diagrams for Cantilevers. 

Sp = load on A p 
= w X 
This is proportional to x, and therefore the shear diagram is 



124 THE STRENGTH OF MATERIALS 

a triangle, the maximum shear occurring at the end b, and 
being equal to wl or W, if W is the total load on the 
cantilever. 

Mj, = moment of load w x about P 

X 
= W X X ^ 

w x^ 



2 
This is proportional to x^, and therefore the B.M. diagram 
will be a parabola with vertex at a. The maximum B.M. 

will be equal to ^ or -^- and occurs at b. 

Case 4. Cantilever with Isolated Load and Uniform 
Load. — ^In this case, as in Case 2, the shear and B.M. diagrams 
are obtained by drawing the separate diagrams in accordance 
with Cases 1 and 3, and then adding them together as shown 
in the figure. 

Case 5. Cantilever with Uniformly Increasing Load. 
— Suppose a cantilever a b carries a load which increases in 
intensity uniformly from the free end a to the fixed end b. 
Fig. 58. This occurs in practice in the case of a vertical wall 
or side of a tank subjected to water pressure. 

Let the intensity of load at unit distance from Ahe w tons 
per foot run, then the intensity at any point p at distance x 
from A will be equal to w x. The intensity of load at b will 
be w I, and the total load equal 

w I , w l^ ^^r 

2xZ=2=W 

S,. = total load to left of p 

x w x^ ^ 

= wx X ^= - 2 

.'. Shear diagram is a parabola with vertex at a^, the 
maximum shear at b being equal to W. 

M,, = moment of load to left of p 
W X^ X w x^ 

= —7^ X 



3 6 



BENDING MOMENTS AND SHEARING FORCES 125 

.'. B.M. diagram is a curve whose ordinates vary as x^, 
such curve being called a parabola of the third order. 



A S O4C) 3 O zO / 




^p 



Fig. 58.— B.M. and Shear Diagrams for Cantilevers {continued). 



The maximum B.M. at b is equal to -- = ^ 

^ 6 3 

The diagrams then come as shown in Fig. 58. 



126 THE STRENGTH OF MATERIALS 

Case 6. Cantilever with Irregui^\r Load System. — 
Graphical Method. — Suppose a number of loads 0,1, 1,2, 
and so on, Fig. 58, act on a cantilever. To obtain the shear 
and B.M. diagrams set down 0, 1 ; 1, 2 ; 2, 3, &c., do^vn a vector 
line 0,5 to represent the forces to some convenient scale, and 
take a pole p at some convenient distance p from the vector 
line 0,5 and join p to each of the points to 5 on the vector 
line. 

Xow across the lines of the forces draw a g parallel to p o ; 
across space 1 draw a h parallel to p 1 ; across space 2 
draw h c parallel to p 2, and so on until the point / is 
reached. 

Then ah c d e f g i^ \h.Q B.M. diagram. 

To obtain the shear diagram, project the pomts 0-5 on the 
vector line across their corresponding spaces, the line through 
the point being draT\7i right across the span, the stepped 
figure thus obtained being the shear diagram. 

Proof. — Consider any point p along the span, and produce 
a b and 6 c to cut the corresponding ordinate p^ p.. of the link 
polygon at b' and c respectively. 

Xow consider the As a p^ b' and p 01. 

They are similar, and as the bases of similar triangles are 
proportional to their heights, we have 

Pi b' _ « Pi 

1)71 ~ p 

.' . p X -p-^b' = 0, I X ciT^ 

But 0. 1 ■ a Pi = moment of force 0, 1 about p. 
.•. p X PjL 6' =^ moment of force 0, 1 about p. 

Similarly it follows that 

p X b' c = moment of force 1, 2 about p, 
and /; x c' P2 = moment of force 2, 3 about p. 

. • . We see that /) x p^ p.2 = p (Pj b' —b'c — c p,) 

= moment of all forces to left of P about r. 

.'. Since p is a constant quality, it follows that the ordin- 



BENDING MOMENTS AND SHEARING FORCES 127 

ates of the link polygon represent the bending moments at 
the corresponding points of the beam. 

Now consider the shear S at p. The t6tal force to the right 
of P is 0, 1 + 1, 2 + 2, 3 = 0, 3, and this is obviously the 
value given on the shear diagram. 

Scales. — In all graphical constructions it is extremely 
important to state clearly the scales to which the various 
quantities are plotted, and to see that such scales are con- 
venient for reading off. 

Let the space scale be 1 in. = a: feet 
and the load scale on the vector line 1 in. =2/ tons 
and let the polar distance be f actual inches. 

Then the scale to which the bending moments can be read 
off is 1 in. = 2^ X X X y ft. tons. 

p should thus be chosen so as to make this a convenient 
round number. 

To take a numerical example, suppose the space scale is 
1 in. = 4 ft. and the load scale is 1 in. = 2 tons, then if j^ 
is taken as 2 J ins. the B.M. scale will be 1 in. = 4 x 2 x 2J 
= 20 ft. tons. 

If p has been taken as 2 ins. the B.M. scale would have 
come 1 in. == 16 ft. tons, which would not be nearly such a 
convenient scale. 

B. Simply Supported Beams. — ^. e. beams simply 
resting on two supports, the loading all being at right angles 
to the length of the beam. Unless it is definitely stated to 
the contrary, we will always take it that the supports are 
at the ends of the beam. 

In simply supported beams the forces acting are the loads 
and the reactions at the supports, the sum of the reactions 
being equal to the total load, and their values being obtained 
by means of moments. As the ends are freely supported, 
there can be no bending moment at either end. 

We will now consider the following standard cases : — 

Case 1. Isolated Load in any Position. — Let a load W 
be supported at a point c on a beam a b (Fig. 59) of span I, 



128 



THE STRENGTH OF MATERIALS 



the distances of the point c from b and a being b and a 
respectively. 

Then to get the reaction R„ at b take moments round A. 

Then R„ x Z = W x a 
W X a 



R„ = 



Similarly R, 



I 

W X 6 
I 



Now consider a point p between b and c. 

+ Wa 



S,. = R„ = 



I 




® 



w A 

2 



R 



'////////. 



y//////A 



AW 
2. 



Sh&ar Diaqnani 




Fig. 59. — Simply Supported Beams. Isolated Load. 



. • . between B and c the shear diagram is a rectangle of 
height = - , 

Now take a point p' between c and a. 
S,, = R„ - W 



w_«_w = w(^^ 



Wb 



I 



= -R. 



• . Shear between c and a is a rectangle of height 

_-Wb 

~ I 



BENDING MOMENTS AND SHEARING FORCES 129 

In the case of the cantilever there was no need to dis- 
tinguish between positive and negative shear because there 
was no change in direction of the shear; but in the present 
case there is a change in direction, and so we will use the 
rule given on p. 121. 

Now considering the bending moment, 

Mp = R„ X a; = 

This is proportional to x, and therefore the B.M. diagram 
between b and c will be a triangle, the B.M. at c being equal 

to — J — . If p were between c and a and at distance x' from 

A we should have 

Mr = R3 (^ - x') ~V^ {I - x' - b) 

= nj - n, . x' - w I + w x' + w b 

= x' (w -n,) -hWb -i{w - R,) 
= n,.x' + wb-in, 

Wbx' 



Wbx' 

I 



+ Wb -Wb 



This is proportional to x\ and therefore the B.M. diagram 
between a and c is also a triangle, the whole diagram then 
coming as shown in the figure. 

Case 2. Isolated Load at Centre. — This is a special 

case of the preceding one, in which « = 6 = - 

W 

Each reaction is now equal to -^- and the maximum 

W X ^ X ^ W? 

Case 3. Uniform Load over Whole Span. — Let a uni- 
form load of w tons per ft. run cover the whole span a b, and 
consider a point c at distance x from b. 

K 



130 



THE STRENGTH OF MATERIALS 



In this case the two reactions will, from symmetry, be equal, 
and each have the value -^ or - - 

Then ^^ — V^,,^ — w x — w {^ — x 

This is a linear relation, therefore the shear diagram will 
be a triangle as shown, having values ± %-^ at the ends and 
changing sign at the centre. 



rtr Torii, p>er Ft 




pnoc^hancp 






^ffi 



a. »■-*- 



K.^^??^ 




Fig. 59o. — Simply Supported Beams. Uniform Load. 



Wa 

if 



Now consider the bending moment. 



X 



M^. = R„ X a; — t^;.T X 

Wlx WX^ _W , 2\ 

^ "2 ~ 2 ^ 2^ ~ ^' 

This depends on x^, and therefore the B.M. diagram will 
be a parabola. 

The maximum B.M. will occur at the centre — i. e. when 
I 



X = 



2' 



BENDING MOMENTS AND SHEARING FORCES 131 

Then maximum B.M. = ^ y^J ~ ^ 2/ "" 2 ' 2 " 4/ 

w P- wP Wl 
2^4=8 ^^T" 

Case 4. Uniform I^dad over Portion of Span. — Let a 
uniform load of w tona per foot run and of length e d equal 
to / be i^laced on a beam a B of span l, and let the centre c 
of the load be at distance a and b respectively from a and b. 

Then, if total load wl = W, 

K,, = ~ and K, = 

The shear between b and d will be constant, and will be 

Wa 

equal to ; between d and E the shear will decrease 

uniformly until at e the shear will be equal to 

R„-W = W«-W=^^* = -R„; 

between e and a the shear will be constant and equal to 

- Wb 

J , the shear diagram then coming as shown on the 

figure. 

The point k at which the shear is zero can be found as 
follows. Let it be at distance x from the centre c of the 
load. 

Then S^ = R„ - w (^ " ^) ^ ^ 

Wa wl , 

I.e. — ^ -[- 'W X ^ 

L 2 

wl W a wl wla 
">-^= 2 - ,, = 2 - T 

•'• ^~ 2 ~ T " ^ V2 ~ L 

The B.xM. diagram can be drawn by setting up a length 

W/ 
^'2 ^'' '^ o > i- e. the bending moment at the centre of the 

o 

short span e d, then produce Cg f to o, making f o equal to 



132 THE STRENGTH OF MATERIALS 

Cg F and join g to Ej and Dg, and produce to meet the reaction 
verticals in Ag and Bo. Join Ag Bg, and we then have the B.M. 
diagram as shoAvn. 

To prove that this gives the correct diagram, consider the 
bending moment at a point at distance x from the centre of 
the load. 

Then M, = R, (6 - .r) - | (| - xj 

Wa ,, , W/i Y n^ 

= -^- (6 - ^-l - 21 (.2 - •■^j W 

Now, °^ = Al« 

G C2 Eg Cg 

GCoXAoQ WZ a Wa 
Eg Cg 4 J^ 2 

Similarly g r = -^ 

.-. QR = ^ (6 — a 

Q R X a W a ,, 

Wa , Wa,, 

.-. GH = GQ + QH = -^^~ + 2Y ^ ^^ 



W a / h — a 



Wa / L + 6-a \_Wa (6 + 6) _ Wa6 
2 V L / 2l l 



. . J p h — X 

Agam = — -^ 

^ GH b 

h — X W a (6 — x) 

.-. JP = — V — X GH = ■ ' 

h L 

, .JO o Do 
Agam = ^ 

G Cg Cg Dg ^ 

G Co X O D, W / 2 

.-.JO = — ^ =-7— X -s — 

Cg Dg 4 Z 



W /7 

T 2 - ^ 



BENDING MOMENTS AND SHEARING FORCES 133 
Then since curve is a parabola — 

FC, — ON /0C2^^ 



F C2 \C2 D2 

Wl 

8 x^ 



Wl ~ l^ 
8 4 

W Z _ _ W Z 4.x^ _ Wjr2 

• • "8 ^^ ~ S ^ l^ ~ 21 
Wl Wx^ 

.•.NP=PJ — JO + ON 

Wa{b -x) W fl ^\ , WZ Wx^ 



L 2 V2 ; ' 8 2Z 

Wa ,, , W fx^ P P Ix 

= --ib-x) --^. ^^-g-+^^--- 
W a ,, , W /x^ Ix l^ 

Wa., . W fP ; , 2 

(0 — a;) — ;^ J — I X + X'^ 



L ' ' 2Z \4 

Wa ,, , W /Z V 

= -IT (^ - ^) - 21 12 - ^; 

Comparing this with (1) we see that N p gives the B.M. at 
the given point. 

We shall prove later (p. 149) that the B.M. is a maximum 
at that point of the span where the shear is zero, and so the 
vertical through k will give the maximum ordinate of the 
B.M. diagram. 

Alternative Construction. — The following alternative 
construction is usually more accurate in practice. On a hori- 
zontal base Ag B2, Fig. 60, set up Cg m equal to , ^. e. the 

Ij 

B.M. due to an isolated load W placed at c, the centre of the 
load. Join m Ag, M B2, cutting the verticals through E.and d in n 
and D respectively, and join n d. On Eg Do draw a parabola 

W I 

Eg Q D2 of height equal to -^ (the B.M. due to a uniform 



134 



THE STRENGTH OF MATERIALS 



load on a span e d), then the B.M. diagram comes as shown 
shaded. If it is desired to have the diagram on a straight 
base, the parabola may be drawn on the inclined base n d as 
indicated in dotted lines. The proof of this construction is 
left as an exercise to the student. 

Case 5. Irregular Load. — Graphical Construction. — 
Let a number of loads Wi, Wg, W3, and W^, be placed knj- 
where along a span ab. Number the spaces between the 
loads and set down, 0, 1 ; 1,2; 2, 3 ; 3, 4, as a vertical vector 

- I . 



ooocooorrxTrr) 




^ 8 

Fig. 60. — ^Alternative Construction for Uniform Load over 
Portion of Span. 



line to represent the loads to some convenient scale, and in 
any position take a point p at convenient polar distance p 
from the vector line, and join p 0, p 1, P 2, etc. 
• Across space then draw a h parallel to p ; across space 1 
draw h c parallel to p 1 and so on until e / is reached, this 
being parallel to P 4. 

Join a f, then the figure a, b, c, d, e, /, a, will give the B.M. 
diagram for the given load system. 

Now draw p x parallel to a /,/the closing link of the link 
polygon then on the vector line, 4 :«: = R„ and xO = R^. 

To draw the shear diagram, draw a horizontal line through 



BENDING MOMENTS AND SHEARING FORCES 135 

X right across the span : this gives the base line for shear. 
Now project the point horizontally across space ; project 
point 1 across space 1 and so on, the stepped diagram thus 
obtained being the shear diagram. 

Proof. — Produce the links cb, d c, e d, f e back to meet the 
vertical through a in b\ c\ d\ e\ and let the first link a b 




Fig. 61. — Graphical Construction for Shear and B.M. Diagrams. 

produced meet the last link e / in y. Then the point Y is 
the point through which the resultant of the loads acts. 
Now the triangles abb' and p 1 are similar. 
. a 6' _ 0, 1 

h ~ V 

0, 1 X Zi Wi X ^1 



.-. aV = 



V P 

moment of first load about a 
- 



136 THE STRENGTH OF IVIATERIALS 

q' *i 1 ^ 7 ' ' _ moment of second load about a 
imi ar ^ c - 

and so on 

_ sum of moments of loads about a 
_ ^_ 

but R„ X L = sum of moments of loads about a 

, R„ X L 

. • . a e = — 

P 

Now consider As a e' f and a; 4 p ; they are similar : 
a e' _ 4 X 
• • IT ^^ 

. p X a e' T^ 

.' . 4:X = ^ = R^ 

L 

Similarly a: = R^ 

Now consider any point r along the span. 
S„ = R3 - W4 

but the ordinate s of the shear diagram is equal to 3 x, and 
therefore the stepped figure gives the correct shearing force at 
any point. 

Let the vertical through r cut the B.M. diagram in R^ Rg 
and / e produced in eg- 

Then by exactly similar reasoning as before 

T5 moment of R„ about r 

Ri e, = — 

P 

^ moment of W 4 about R 

Kg 63 = 

P 

. • . Rj R2 = R^ ^2 — 1^2 ^2 

_ moment of R^ — moment of W 4 about r 
~ P 

P 
. • . M„ = ^ X Rj R2 

. * . The ordinate of the B.M. diagram represents the B.M. 
at any point. 



BENDING MOMENTS AND SHEARING FORCES 137 

Scales. — As in the case of the cantilever (p. 126), if 
r' = .T feet is the space scale and V ^ y tons is the force 
scale, and if the polar distance is p actual inches, then the 
vertical ordinates of the B.M. diagram represent the bending 
moment to a scale V^ = p x x x y it. tons. 

Note. — In this construction the bending moment R^ Rg 
is measured vertically and not at right angles to the closing 
line a f. 

Case 6. Irregular Load — Overhanging Ends. — The 




Fig. 62. — Beam with Overhanging Ends — Graphical Construction. 



construction just described is equally applicable to the case 
where the ends are overhanging. Fig. 62 shows such a case. 
Set out the loads down a vector line as before and take any 
pole p. Now draw a b parallel to p across space 0, i.e. 
between the support vertical and the first force line. Then 
draw h c parallel to p 1 across sj^ace 1 and so on, the last link 
e / being drawn between the last force line and the reaction 
vertical. Joining a f we get the B.M. diagram as shown. 

To get the shear diagram draw p 5 parallel to a /, then the 
horizontal through gives the base line for the shear between 



138 



THE STRENGTH OF MATERIALS 



A and B. The shears in the end spaces will be equal to the 
end forces 0, 1 and 3, 4 respectively, as shown on the figure. 

This graphical loading is applicable to all kinds of loading, 
and any of the previous standard cases can be worked by its 
means. In the case of a continuous load the latter should 
be divided up into a number of small portions, and the load 
in each portion treated as an isolated load acting down the 
centre of such portion. 




Parabola of 3''4 Order 



JB, 



Fig, 63. — Simply supported Beam with Uniformly Increasing Load. 

Case 7. Uniformly Increasing Load. — Suppose a beam 
A B carries a load which increases in intensity uniformly from 
the end b to the end a. Let the intensity of the load at unit 
distance from b he w tons per ft. run ; then the intensit}^ at 
any point p at distance x from b will be equal to w x (Fig. 63). 

The intensity of the load at a will be equal to w I, and the 

I w Z^ 
total load W will be equal to w I x ^ = -^ 



BENDING MOMENTS AND SHEARING FORCES 139 
The resultant load W acts through the centroid of the load 
curve, i.e. at distance from a. 

W 

R. = -3 

Then Sp = total load to right 
_ W w x'^ 
~ 3 2~ 

This depends on x^ and therefore the shear curve is a parabola. 
The point c^ is obtained as follows — 

S/ 



= 


W w x^ 
~ 3 2 




' x^ 


W wP 




2~ 


3 2x3 




x^ 


P 
~ 3 




X 


= "L = -577 I 
V3 




Mp 


= R3 X .T - 2 

W .-r w x^ 


' 3 



3 6 

This depends on x^, and so the B.M. curve is a parabola of the 
third order. 

The maximum B.M. occurs at the point of zero shear (see 

p. 149), i.e. when x = ,- 

V3 

. . Maximum B.M. = >— — , 

3V3 18V3 

1 1 \ 



>,3V3 9V3/ 
_ 2W? _ 2WZa/3 
~9\/3^ 27 
= 128 WZ. 

The B.M. and shear curves then come as shown in the figure. 



140 



THE STRENGTH OF MATERIALS 



Case 8. Uniformly Loaded Beam with both Ends 
Overhanging.^ — Let a beam of span l be loaded with a uniform 
load of w tons per foot run, and let it overhang a distance 
X at each end, the distance between the supports being I. 

The overhanging portions act as cantilevers, and the shear 
and B.M. diagrams for such portions will be as shown. The 
B.M. for the centre portion will be a parabola drawn on the 
base shown dotted, the resulting curve being as shown cross- 
hatched. 

urfcns per Ft 

nooooonoocYYTXTmoono. 



JC 



^^z^. 



I 

L 



Sh 



ear 



B.M. 




X V 



Vol 



Fig. 63a. — ^Uniformly Loaded Beam with Overhanging Ends. 



If the load on the centre portion of the span were removed, 
the B.M. diagram would consist of the two end parabolas 
and the dotted line. This B.M. is opposite in direction to 
that due to the centre portion, and therefore on replacing the 
centre load and drawing the parabola, the resulting curve is 
the difference between the two as shown. 

To find the value of x to get the least resultant B.M. w^e 
proceed as follows. 

As X increases, the B.M. at the supports increases and the 
resulting B.M. at the centre decreases, so that the least 
B.M. will occur when the svipport B.M. is equal to the centre 
B.M. 



BENDING MOMENTS AND SHEARING FORCES 141 

The support B.M. = ^-^ 

The centre B.M, = ^f^ - ^^- 

Tn , 1 , W X^ WP W X^ 

It these are equal —^r- = -^ ^- 

Z o Z 

„ wP- 
. • . w x^ — —^ 

o 
I 

^ ~ 2 V"2 



L I -\- 2x J I 
1 2 



. V2 2 + V2 

^+2 

^2J^-V2)^2-V2 = -586 
z 

This gives the position at which the legs of a trestle table 
should be placed to give the maximum strength to the latter. 

Case 9. Uniformly Loaded Beam with One End 
Overhanging. — A beam a b, Fig. 64, of span l is supported 
at one end a and overhangs the support c at the other end ; 
we wish to find, with a uniformly distributed load, the position 
of the support c which will be most economical — i. e. give the 
least bending moment. 

If the length of the overhanging portion B c is /g ^^^ the 

distance a c is l^, the B.M. diagram will be as shown shaded in 

the figure ; the portion b^ d is a parabola tangential at b^ and 

is the familiar diagram for a cantilever with a uniformly dis- 

to I 
tributed load ; the portion a G c^ is a parabola of height — ^ , 

o 

the usual one for a freely supj)orted span a c ; and a^ d is a 
straight line. 

The maximum positive B.M. will be given by kj, which 

on (J Cl\ 

will be equal to — "^ q- since the B.M. between the point 



142 



THE STRENGTH OF MATERIALS 



Ai and the point e of contraflexure will be the same as for 
a freely supported beam of span a^ e, and the maximum 
negative value is given by CiD. Our problem resolves itself 
into find the position of c to make J k or c^ d the least 
j)ossible. 

Now, if 3"ou move the point c to the left, c^ d will increase 



IC joer unit lenqTh 



A 



L 



wiCj -a) ^ 




I 



Fig. 64. 

and K J will decrease, whereas if c moves to the right the 
converse happens. If, therefore, c^ d = k j, movement of 
c will increase one or the other, so that the least value of 
either occurs when they are equal. 



This gives 



8 2 

i.e. (/i - af = 4/2'- 



or [li — a) 



21.-. 



(1) 



BENDING MOMENTS AND SHEARING FORCES 143 

Again, by the property of the parabola 

---("i^-f) (2) 

This can be found by taking the B.M. at e for the span 
Ai Ci ; also by similar As. 

E F _ Aj E 
Ci D Ai Ci 

I.e. E F = -^ X 11- (3) 

Combining (2) and (3) we get 

or, a = -? (4) 

h 

Putting this result in (1) we get 

n J~ ^^ ^ ^2 

i.e. l^^ -21^1^-1^^ ^0 (5) 

The solution of this quadratic equation gives, taking the 
positive root — 

... 1 + ^^ = ^^4-^2 _ L J ^ 2-^^^ _ 3.^^^ 

^2 ^2 ^2 

or, ?o = s— 7TT *• 6. ?o = '293 L 
^ 3414 

In this case the maximum B.M. will be equal to 
ivl^ _ w X (-293 L)^ wj.^ 
2 " 2 ~ 23-3 

It will be of some interest to compare this result with that 
which would occur if each end were overhung and the sup- 
ports were placed so as to give the least B.M. for this condition. 

In this case the best condition is given when the overhang 
is '207 L. This gives a maximum B.M. equal to 

w X (-207 L)2 w l2 

2 = 46^6 ^PP^'°'^' 

which is half that for the previous case. 



144 THE STRENGTH OF MATERIALS 

Steps in Shear Curves. — In practice it is impossible to 
get absolutely sharp steps in shear diagrams, because the 
load cannot be transmitted at a mathematical point, but 
must be distributed over a short length. This has the effect 
of slightly rounding off the corners of the shear diagram as 
shown exaggerated in dotted lines on Fig. 65a. 

Numerical Examples. — (1)^ freely supported beam of 
20 ft. span carries a uniformly distributed load of 5 tons, and 
isolated loads of 3 and 2 tons, at distances respectively of 4 and 
5 ft. from the ends {see Fig. 65). 

We have first to get the reactions R^ and R„. 
Take moments round b. 

R, x20-5xl0 + 3xl6 + 2x5 
= 50 + 48 + 10 = 108 
T> 108 . . ^ 

.-. R„ = 10 - 5-4 = 4-6 tons. 

The shear diagram then comes as shown in the figure, the 
amounts of the steps being equal to the isolated loads. The 
point at which the shear is nothing is found as follows — 

Let it be at distance x from b. Then 

S, = = R„ - 2 - i(; . X 

= 46-2- ^^ 
20 

= 2-6 - ^, 
4 

.-. ^ = 2-6 
4 

X = 10-4 feet. 

The B.M. at this point will be a maximum, and will be 

equal to 

1 10-42 
M X = R,. X 10-4 - 2 (10-4 - 5) - ^ . —^ 

= 47-84 - 10-8 - 13-52 
- 23-52 ft. tons. 
The B.M. diagram will consist of a parabola for the uni- 
formly distributed load, the max. ordinate of which is equal 



BENDING MOMENTS AND SHEARING FORCES 145 

5 X 20 



to 



8 



= 12*5 ft. tons. The B.M. diagram for each of 

the isolated loads will be a triangle, the respective heights 

3 X 4 X 16 (.a t4. 4. , 2 X 5 X 15 „ 

being ^^ = 9'6 ft. tons, and ^a = '^ *^- 

tons. Combining these three figures we get the B.M. diagram 

.sTons 2 Tons 



^'Fb 




4 -67 



Fig. G5. 

■shown on the figure, and on scaling off the maximum ordinate 
it will be found to be 23*5 tons. 

Note. — In all constructions where diagrams are going to 
be added together, such diagrams must of course be drawn 
to the same scale. 

(2) A girder oj 24 ft. span is supported at one end, and rests 
on a column at a point 6 jt. from the other end. The girder 
carries a uniformly distributed load of 6 tons and an isolated 
load of 2 tons at the free end. Draw the shear and B.M . curves. 



146 THE STRENGTH OF MATERIALS 



/? 




A, 



6 Tons distributed 



^To 



ons 




P4 3Pf\^ . 



E> 




3.M.j^or Isolated Load 




E> M for Unijorm Load 




C. 



Combined B.M. 
Fig. 65a. 

To jfind the reactions take moments round a (Fig. 65a), 

Then 

18 R„ = 6 X 12 + 2 X 24 = 120 

•"• "^" "" 18^ ^ ^^ ^^^^ 
.-. R^ = 8 - 6f = 11 tons. 



BENDING MOMENTS AND SHEARING FORCES 147 

The shear at c will be = 2 tons. It then increases until the 
point B is reached, when its value becomes equal to 35 tons. 
It then suddenly changes sign to a value 3*17 tons and then 
decreases uniformly to the end A, where the value comes 1'33. 
The shear diagram then curves as shown in the figure, the 
dotted lines indicating what occurs in practice owing to the 
impossibility of getting the loads and reactions concentrated 
on a mathematical point. 

Considering first the B.M. for the isolated and uniform 

loads separately, the B.M. curve due to the isolated load will 

come as shown in the figure, the B.M. at b being equal to 6 x 

2 = 12 ft. tons. Now, considering the uniform load, the 

diagram for the portion b c will be a parabola with vertex 

wP" 1 6^ 
at c, the ordinate b^ d at B^ being = = -^ x ^ = 4'5 ft. 

tons. Then between b and a the B.M. curve due to this 
overhanging load will be the straight line Aj d, as such over- 
hanging load requires an isolated balancing load at a. 

The B.M. curve for the portion a b will be a parabola of 

wV^ 1 18^ 
central height = —^ = v x „- = 1012 ft. tons, the shaded 

portion being the resulting curve for the central and over- 
hanging portions of the uniform load. Combining these 
diagrams we get the resulting B.M. curve as shown, the 
max. B.M. occurring at B, and being equal to 165 ft. tons. 

Relation between Load, Shear and B.M. Diagrams. 
— Let A c' T>' B, Fig. 66, represent the load curve on a span a b. 
Take any point p along the span, and consider a short 
piece c D of the load, the centre of which is at distance x 
from p. 

Then the shear at p due to this piece of the load will be equal 
to the area of the portion c d of the load curve. Therefore the 
total shear S,. at p will be equal to the area of the load curve 
up to that point. 

But a sum curve * is such that its ordinate at any point 

* See p. 162. 



148 



THE STRENGTH OF MATERIALS 



represents the area of the primitive curve up to that point. 
Therefore the shear curve is the sum curve of the load curve. 

Suppose b' F E G a' is the sum curve of the load curve. 
Now consider the B.M. at p. 




Fig. 66. — Relation between Load, Shear and B.M. Diagrams. 



The B.M. at p due to the portion c d of the load 
= given portion of load x x. 

Now if E and r are the corresponding points on the shear 
curve, the difference of the ordinates at E and f gives the 
load on the portion c d. 

. • . Load on portion c d = e f^. 

. • . B.M. at p due to portion c d = e f^ x a:. 

.' . Shaded portion e f Fg e^ represents the B.M. at p due 
to the portion c d of the load. 



BENDING MOMENTS AND SHEARING FORCES 149 

.• . Total B.M. at p = M,. = area of shear diagram up to p. 

Thus the B.M. curve is the sum curve of the shear curve. 

So that by drawing the sum curve B J H of the shear curve 
we get the B.M. curve. 

Scales. — If V ^ x tons per foot is the scale of the load 
curve, and p-^ is the polar distance measured on the space 
scale for obtaining the shear curve, then the scale of the 
shear curve V "^ Pi ^ tons. If P2 is the polar distance from 
which the B.M. curve is obtained, measured on the space 
scale, the B.M. scale will be V^ = PiPi^ foot tons. 

Point of Maximum B.M. — If the B.M. is a maximum, the 
tangent to the curve at this maximum must be horizontal, 
and therefore the corresponding ordinate on the shear diagram 
must be zero in order for the line through the pole to be also 
horizontal. 

Thus we get the rule that the maximum B.M. occurs where 
the shear is zero. 

The base lines s s and m m of the shear and B.M. curves 
depend on the manner in which the ends are fixed. If one 
end is free, the shear and B.M. at this point are zero. If 
one end is freely supported the shear at this point will be 
equal to the reaction, and the B.M. will be zero. 

The above relations are expressed mathematically as 
follows : Let the load at any point at distance x from the 
origin be F {x) 

Then the shear at the point will be = / ¥ {x) d x -{- c^ and 

the B.M. will be =yyF {x) d x + c^x -{- c^. 

The integration constants c^ and Cg depend on the manner 
in which the ends are fixed, and correspond to the base lines 
above referred to. 

A Template for Bending Moment Diagrams. — For 
Various Cases of Uniformly Distributed Loads. — In 
designing beams carrying uniform loads it is necessary in 
order to draw the Bending Moment diagrams to draw para- 
bolas ; the usual procedure is to draw the parabolas for the 



150 



THE STRENGTH OF >L\TERIALS 



special arrangement of the loads and for the particular 
manner in which the beams are supported, this involving 
a good deal of geometrical construction. A temj^late can, 
however, be used to serve for a large number of cases in 
the following manner — 

On a base a b, Fig. 67 — for convenience say 5 ins. long — 




Fig. 67. 



draw by the usual construction a parabola a d c with vertex 
at A, the height b c being for convenience equal to a b. 

A template of the form a b c d can then be made, a 45° 
set-square being a convenient form to cut it from. A pro- 
jection is preferably j^rovided as shown to avoid a sharp point 
which is liable to break off. By means of this template and 
a suitable choice of scales, the Bending Moment (B.M.) 
diagrams for a large number of cases can be then drawn as 
follows — 



BENDING MOMENTS AND SHEARING FORCES 151 

Case 1. Cantilever fully Loaded. — Draw the span e f, 
Fig. 68, to a suitable scale, so that e f is not larger than a b ; 
erect a vertical at the fixed end f and place the template on 
the paper with the point A coinciding with the free end e and 
draw in the curve to the point g where it meets the vertical 
through B. The B.M. diagram is then as shown shaded. 

Scales. — If the intensity of the load is w lbs. per foot run, then 
the B.M. scale will be the square of the space scale multiplied 



by 



Take for instance the case where the space scale is 




Fig. 68. 



1 in. = 2 ft. and the load is 1000 lbs. per foot run ; then B.M 
4 X 5 X 1000 



scale is \" 



10,000 ft. lbs. 



Case 2. Simply-supported Beam fully Loaded. — In 
this case, Fig. 69, we draw e^ f^ to represent the span and 
we draw as before k vertical f^ g^ at one end; the template 
is then placed on the paper with the point a coinciding with 
the point e^ at the other end of the span and the curve is 
drawn until the vertical is cut to the point G^. Now join 
Gi Ej, the B.M. diagram then coming as shown shaded, the 
Bending Moment at any point h being found by projecting 
vertically and measuring the height x as shown. 

The scales are obtained as previously described. 



152 THE STRENGTH OF iLlTERIALS 



E. 




Fic. 69. 




Fig. 70. 



BENDING MOMENTS AND SHEARING FORCES 153 

Case 3. Uniformly-loaded Beam Overhanging the 
Support at one End. — In this case, Fig. 70, we place the 
template on the paper with the vertex of the parabola at the 
overhanging end Fg and draw in the curve until we meet at 
L the vertical through the other end Eg; then join l to the 
other support point k, the shaded area giving the B.M. 
diagram, the Bending Moment at any point being read off 
by projecting vertically as explained in the previous case. 




Fia. 71. 



At points such as m where the B.M. diagram crosses itself, 
the Bending Moment changes sign ; this of course corresponds 
to a reversal of the tension and compression flanges of the 
beam. 

Case 4. Uniformly-loaded Beam Overhanging at 
EACH End. — To obtain the B.M. diagram in this case with 
the aid of the template, we place the template on the paper 
with the vertex of the parabola coinciding with one end E3 
(Fig. 71) and draw the curve until we meet the vertical through 
the other end at q. 



154 



THE STRENGTH OF ^UTERIALS 



Join Q to the mid-point r of the span, the line Q r cutting 
the vertical through the support p at the point s. Einally 
join s to the other support point N", the B.M. diagram then 
coming as shown shaded in the figure. 

Case 5. Uniformly-loaded CoNTrs'Uors Beam of Two 
Equal Span's. — We can get the B.M. diagram in this case 
with the aid of the template by placing it on the paper with 
the vertex coinciding with the end support E4 and drawing 
in the curve mitil it intersects at the point G3 the vertical 
through the centre support G3 ; then by reversmg the template 
and commencing the curve at the other outside support E5 
we shall get the reversed curve going from G3 to E5 as shown 
in Fig. 72. 




Fig. 



A length G3 t is then set down from G3 of length equal to 
J G3 F4 and, by joining t to E5 and E4, we get the B.M. diagram 
as shown shaded in the figure. The student should check 
the correctness of this method after reading Chap. XV. 

The scales are obtained as previously explained. 

Numerical Exa^iple. — Take a continuous beam of two equal 
spans each 16 ft. long, each span being covered by a load of 
1500 lbs. per foot run. 

Taking a linear scale of T' = 4 ft.. E4 F4 will be 4 ins. 

Then the B.M. scale will be, as explained above, 



42 X 5 w; 16 X 5 X 1500 



1" = - \~ = 



= 60,000 ft. lbs. 



If the distance Gq t be measured, it will be found to come 



BENDING MOMENTS AND SHEARING FORCES 155 

equal to '8 inch with a template of the dimensions suggested 
in Fig. 67. 

.-. Maximum B.M. = '8 x 60,000 = 48,000ft. lbs. 

The B.M. at any other point u can be obtained by reading 
off the vertical ordinate x to this scale. 

In the case of a beam supported freely at one end and 
securely fixed at the other, the B.M. diagram will come the 
same as one-half of that shown in Fig. 72, the point E4 being 
the freely-supported end and the point F4 the fixed end. 

A number of other, cases might be given, but we think 
that the above are sufficient to show that a template of this 
kind would be of considerable assistance to draughtsmen for 
obtaining the B.M. diagrams for a variety of cases. 

In some respects the template would be more easy to make 
if it were made of the shape a d c b. Fig. 67, because the con- 
vex curve can be somewhat more readily shaped. If, instead 
of the given dimensions for a b and B c, other values are taken, 
the rule for scales must be correspondingly amended, bearing 

A B^ 

in mind that B c should represent ^y— for the B.M. scale to 

be equal to the square of the space scale when w ^ 1. If b c 
has not this value, then the B.M. scale will vary in the inverse 
ratio. 

B.M. AND SHEAR DIAGRAMS FOR INCLINED LOADS 

In all the cases that we have considered up to the present 
all the loading has been at right angles to the length of the 
beam. We will now consider some cases in which this is not 
the case, and will take both horizontal beams with non- 
vertical loads and sloping beams. The principal difference in 
this case is that there will be thrust in the direction of the 
beam, and we shall have a curve of thrust in addition to 
the curves of shear and B.M. 

The general rule is to resolve all forces, including the 
reactions, along and perpendicular to the beam. From the 
forces along the beam a curve of thrusts can be dra^vn, and 



156 



THE STRENGTH OF MATERIALS 



from the forces perpendicular to the beam the curves of 
shear and bending moment are drawn in the ordinary manner. 
We will define the thrust at any point of a beam as the 
sum of the components in the direction of the beam of all 
the forces to the right of it, remembering that if the thrust 
is negative it becomes a pull. 




Thrust 0>iacjram 
Fig. 73. — Beam with Inclined Loads. 



Case 1. Horizontal Beam Freely Supported sub- 
jected TO Inclined Loads. — Let a beam a b have inclined 
forces Fi and F2 (Fig. 73) meeting the centre line in c and D. 
Let the end a rest on a free support and let the end b be freely 
supported, but prevented from longitudinal movement as 
shown. If the resultant of F^ and Fg acted towards the end 
A, then this end would have to be prevented from movement. 



BENDING MOMENTS AND SHEARING FORCES 157 

Resolve the forces F^ and Fg into vertical and horizontal 
components W^ Wg and Q^ Q2 respectively. 

Then R„ will be inclined, the vertical component Wi, being 
that found by considering the forces W^ W2 in the ordinary 
way and the horizontal component Q^ being equal to Q^ + Qg. 

The reaction R^ will be vertical, and will be obtained by 
considering the forces W^ and Wg in the ordinary way. 

If the resultant of F^ and Fg were found it would pass 
through the intersection of R^ and R^, since three forces in 
equilibrium must pass through a point. 

The shear and B.M. diagrams are then found in the usual 
way for weights W^ and Wg, and are as shown. 

The thrust diagram is obtained by plotting up at each 
point the value of the thrust, and this comes as shown. The 
same method applies for any number of loads, two having 
been chosen to give simplicity of figure. 

Case 2. Inclined Beam with Vertical Loads. — Re- 
actions Parallel. — Let an inclined beam a b (Fig. 74) be 
supported freely at a and pin- jointed at b. Then if it be 
subjected to vertical forces F^ and Fg at c and d, the reaction 
at A, and therefore also that at b, must be vertical, their 
values being found in the ordinary manner. 

Now resolve the weights and reactions along and per- 
pendicular to the beam, obtaining weights W^, W^, W2, Wa? 
and thrusts Q3, Q^, Q2, Q.,- 

Then the B.M. diagram can be drawn either on a sloping 
base A B or the projected horizontal base A^ b^. 

M„ = W« X D b 

1 D B R„ 

. • . Wb X D B = Rh Zi 

. ' . We see that for a sloping beam with vertical reactions 
the B.M. diagram is the same as for a horizontal beam of the 
same span as the horizontally projected length of the sloping 
beam. 



158 



THE STRENGTH OF MATERIALS 



The B.M. at a point p, for example, is obtained by drawing 
a vertical through it, a b representing the B.M. 

The shear and thrust diagrams are obtained as shown, and 
will be easily followed from the figure. 




Fig. 74. — Inclined Beam with Lower End freely supported. 



Case 3. Inclined Beam with Vertical Loads — Top 
Reaction Horizontal. — In this case the resultant load must 
first be found. Let this resultant act down the line x x 
(Fig. 75). The reaction R^, at B must be horizontal, so draw 
B X horizontal, then if this meets the line x x, R,v must also 



BENDING MOMENTS AND SHEARING FORCES 159 

pass through x, so that by joining a x we get the direction 
of R,. The values of R, and R^, are then found by a triangle 
of forces a, b, c. 

Now resolve the weights and reactions as before along and 
perpendicular to a b. The perpendicular components will be 




Thrust Oiaaram 
Fig. 75. — Inclined Beam with Top End freely supported. 

the same as before, and so the B.M. and shear diagrams will 
be the same as in the previous case (Fig. 74). 

The thrusts will be different, and will be as shown on the 
figure, which will be clearly followed. 

Case 4. Sloping Cantilever. — This is worked in a similar 
manner. Consider, for example, a uniform load of intensity 
m; on a cantilever of length I at an inclination (Fig. 76). The 



160 



THE STRENGTH OF IMATERIALS 



B.M. curve will be a parabola. Its maximum ordinate will be 

iv I cos 6 

rt J because the total weight will be w I, and it acts at 

a distance ^ from the abutment. The shear diagram 




CiT 



Fig. 76. 



will be a sloping straight line, the maximum shear being w I 
cos 6; the thrust diagram will also be a sloping straight line, 
the maximum thrust being w I sin 0. 



CHAPTER VI 

GEOMETRICAL PROPERTIES OF SECTIONS — AREA, 
CENTROID, MOMENT OF INERTIA, AND RADIUS 
OF GYRATION 

Before considering the relation between the Bending 
Moment and the stresses in a beam, we will consider some 
geometrical properties of sections which, as we shall find 
later, are involved in that relation. 

The Determination of Areas. — (a) Mathematical 
Method. — If F [x) represents a function of x and the graph 
of the function be drawn, then the area between graph and 
the axis of x is given by the expression 



A =y'F{x)dx 



In practice, in the determination of areas, this method may- 
become practically unworkable if the equation of the curve 
cannot be simply expressed or if the integration cannot be 
performed. When these conditions occur we have to rely 
on the planimeter or on the following. 

(b) Graphical Method. — If a curve be plotted on a hori- 
zontal base and a new curve be drawn, such that its ordinate 
at any point represents the area of the given curve up to that 
point, the new curve is called the Sum curve or Integral 
curve of the given curve, which is called the Primiitive curve. 

The sum curve can be obtained graphically as follows : Let 

A c D, Fig. 77, be any primitive curve on a straight base A B. 

Divide a b into any number of parts, not necessarily equal (but 

for convenience of working they are generally taken as equal). 

These so-called base elements should be taken so small that 
M 161 



162 



THE STRENGTH OF MATERIALS 



the portion of the curve above them may be taken as a straight 
line. About 1 cm. or '4 in. will usually be a suitable size and 
in most cases a smaller element 11 will come at the end. Find 
the mid-points, 1, 2, 3, etc., of each of the base elements and 
let the verticals through these mid-points meet the curve in 
1 a, 2 a, 3 a, etc. Now project the points on to a vertical 
line A E, thus obtaining the points I b, 2 b, Z b, etc., and join 
such points to a pole p on a b produced and at some con- 
venient distance p from a. Across space 1 then draw a d 




Fig. 77. — Sum Curve Construction. 

parallel to p 1 6 ; d e across space 2 parallel to p 2 6, and so on, 
until the point n is reached. Then the curve a d e . . . n is 
the sum curve of the given curve, and to some scale b ?i 
represents the area of the whole curve. 

Proof. — Consider one of the elements, say 4, and draw / o 
horizontally. 

Now ilf,g,o is similar to the A p, 4 6, a 

^o_46, A 
* / o ~ P A 
but FA = p and 4 6, A = 4, 4 a 
/ o X 4, 4 a area of element 4 of curve 



go 



p 



p 



GEOMETRICAL PROPERTIES OF SECTIONS 163 

^. ., , , area of element 3 of curve , 
Similarlv j q = and so on 

. ' . Ordinate through g^go+fq-{-... 

area of first four elements of curve 
- ^ 

. • . The curve a d e . . . ?^ is the sum curve required. 

Then ii b n he measured on the vertical scale and p be 
measured on the horizontal scale, the area of the whole curve 
will be equal to ^^ x b n. 

It is obviously advisable to make p some convenient round 
number of units. 

The sum curve obtained by this method may have the same 



A 



/}• 



r 


c 


L___ 


< 

, J 

1 J 


a 


' 1 


r-1 


• 








^ i 


* 1 


I i 


i i 


/ 




















\ 


' 


' 


1 


' 


•( 


' ] 


' \ 


1 


f } 


' 1 



B 



B' 



Fig. 78. 



operation performed on it, and thus the second sum curve of 
the primitive curve is obtained, and so on. 

If the operation be performed on a rectangle, the sum curve 
will obviously come a sloping straight line, and if the sum 
curve of a sloping straight line be drawn, it will be found to 
be a parabola. In the case in which it is required to apply 
this construction to a curve which is not on a straight 
base, the curve is first brought to a straight base as 
follows — 

Suppose A cB d, Fig. 78, is a closed curve. Draw verticals 
through A B to meet a horizontal base a' b'. Divide the curve 
into a number of segments by vertical lines at short distances 



164 



THE STRENGTH OF MATERIALS 



apart, and set up from the base a b lengths a^, h^^, etc., equal 
to the vertical portions a, h, etc., on the curve. Joining up 
the points thus obtained we get the corresponding curve 
a' Ci b', on a straight base. 

(c) Simpson's Rule. — Divide the base into an even number 
of equal parts (each equal to c) and measure all the corre- 
sponding ordinates. 

Then area of curve is equal to 



twice sum of four times sum of _ sum of first \ 
"^ odd ordinates and last ordinates J 



3 (even ordinates 




Fig. 79. 



J AT 
-First Moment of an Area. 



X 



{d) Parmontier's Rule. — Divide up base and measure 
ordinates as above, then area of curve is equal to 



Op sum of odd _ ^ f/ second _ first 

ordinates 6(^\ordinate ordinate 



last _ preceding yi 
ordinate ordinate /J 



First Moment of an Area. — Let a small element of area 
a of any figure be situated at the point p. Fig. 79, and let x x 
be any straight line or axis. Then if p n is drawn j^erpen- 
dicular to x x, a x P N is the first moment of the element of 
area about the given line. Now, if the whole figure is divided 
up into elements of area such as a, and the moments of each 
element be taken about x x and the whole of these moments 



GEOMETRICAL PROPERTIES OF SECTIONS 165 

be added together, the resulting sum is called the first moment 
of the area. 

. ' . The first moment of the whole area is the sum of 
quantities such as a x P N. This is expressed symbolically as 
follows — 

First moment of whole area = 'X {a x p n). 

Now the centroid or the first moment centre of an area is 
defined as the point at which the whole area can be con- 
sidered concentrated, in order that its moment about any 
given line will be equal to the first moment of the area about 
the same line. 

Thus if c is the centroid of the area, and c J is drawn 
perpendicular to x x, and the area of the whole figure is A, 
we have 

A + cj = :S(a.PN) 

. • . c J = — ^^ — T -' 

A 

This will not determine the exact position of c, but only its 
distance from the given line x x. If the exact position of the 
centroid is required we must also take moments about some 
other line, not parallel to x x, then the distance from the two 
lines will determine its position. 

In connection with the centroid it should be noted that 
the position of the centroid depends solely on the shape of 
the figure, and not on the position of the axes about which 
moments are taken. As in the case of forces, we have positive 
and negative moments in areas, the moment being positive 
when the given element of area is above or to the right of the 
given axis, and negative when it is below or to the left. 

First Moment about Line through Centroid. — Now 
consider the first moment of an area about a line c c, Fig. 80, 
through the centroid. The moments of elements of area above 
the line such as that at P will be positive, and the moments 
of elements of area below the line such as that at p' will be 
negative. 

Now in this case c J is zero, and therefore A x c j will also 



166 



THE STRENGTH OF :\IATERIALS 



be zero, and therefore we have the rule that the first moment 
of any area about a line tlirough its centroid is zero. 




PosiTio:?^ OF Centroid with Axes of Symmetry. — Suppose 
an area has an axis of s}Tnmetry Y Y, Fig. 81. Then this line 

y 




Fig. 81. 



di^^des the area into two exactly similar halves so that 
corresponding to each element of area at p having a positive 



GEOMETRICAL PROPERTIES OF SECTIONS 167 

moment about Y y we have an equal element at p' having an 
equal negative moment about y y so that the total moment of 
the area about Y Y is zero, or Y Y passes through the centroid. 

If the figure has another axis of symmetry x x, the centroid 
also lies on this line, or we have the rule that the centroid of a 
figure is at the intersection of two axes of symmetry. 

For the determination of the position of the centroid for 
various cases see p. 175-189. 

It should be noted that the centroid of an area is the same 
as the centre of gravity of a template of the same shape as the 
area. 

Second Moments or Moments of Inertia. — The 
rforce(/) ^ 
product of a 4 "^^^^ (^) I by the square of its distance r from a 
[volume {v)j fforce ^ 

given point or axis is called the second moment of the - ^^^^ 
about the given line or axis. [ volume j 

Now, in considering rotating bodies the second moment of 
the mass has to be considered, and this quantity has been 
given the name of the moment of inertia. In the application 
of the second moment to the strength of materials we shall 
have nothing to do with inertia, but the term moment of inertia 
has been generally adopted, and so we shall use it; but we 
must remember that it is really a borrowed term and quite 
an unsuitable one. 

Application to Areas. — If an element of area a is situated 
at the point p, Fig. 82, and p n is drawn perpendicular to a line 
X X, then the second moment of this element of area about the 
line X X is equal to a x p n^. If, as in the case of the first 
moment, we divide the whole area up into elements and take 
the second moment of each, we see that the second moment of 
the whole area about x x is the sum of the second moments of 
the elements. The letter I is always used to denote the 
second moment, the line x x, about which the moments are 
taken, being indicated by writing it I^^. 

Thus we see I^^ = S (a x p n^). 



168 



THE STRENGTH OF ]\L\TERIALS 



In the same way, considering the line y y, we have 

Now suppose K is such a point that the whole area can be 
considered concentrated there so as to give the same second 
moments about x x and Y Y as the second moment of the 
area about these lines. 

Then A x K q2 = I,, 
and A X K R- = I,,. 

Then the x^oint k by analogy might be called the secondroid 




Fig. 82. — Second Moment or Moment of Inertia of an Area. 

of the area with regard to the axes x x and Y Y. The point 
of importance ^vith regard to the secondroid is that its position 
depends on the j)osition of the lines about which the moments 
are taken, whereas the position of the centroid does not. 



RADIUS OF GYRATION 
Now. the distances of the secondroid from the lines x x and 



GEOMETRICAL PROPERTIES OF SECTIONS 169 

Y Y are called the second moment radii or radii of gyration 

about the given lines, and are written k^,. and k,j respectively. 

.'. We have 

A y^,2 = I,.^ = 2 (a X p n2) 



or k,. = ,, ^ 

A kj" = I,, - S (a X P m2) 

or k, = y "2- 

Now, in practice it is nearly always the second moment 
about a line through the centroid that is required, and this is 
obtained as follows — 

Given the Second Moment or Moment of Inertia of 
AN Area about a given Line, to find it about a Parallel 
Line through the Centroid. 

Suppose we know I^x- 

Now, I,^ = 2 (a X P n2) 

= :S [a X (p s + s N)2] = :s [a X (P S + d,,)^] 

= :S [a . (P S2 + 2 P S . 6^, + ^,2)] 

= :S (a . P S2) + ^ (a . 2 P S . 6?,) + S (a . ^/) 

Of the terms on the right-hand side 

S (a X P s)2 = I^.^ (which is required) 
2 (a . 2 p s . d,) = 2 d^, 2 a . p s 

= 2 d,. (first moment of area about line c^ c^ 

through centroid) 
= 2 d_, X 
= 

^ (a . d/) = d/ 5 a 

= d_,^ (area of whole figure) 
= d/.A 
. • . We have I^^ = I,^ + A d/ 
or I„ = I,^ - A ^,2 
Similarly I,.,. = I,.,. — A d,/ 

* The Momental Ellipse or Ellipse of Inertia. — The 

principal axes of a section are defined as two axes at right 



170 



THE STRENGTH OF MATERIALS 



angles through the centroid, such that the sum of quantities 
such as a X p M, p N, or the 'product moment as it is called, is 
equal to zero. 

In the case of sections with an axis of symmetry, such axis 
determines one of the principal axes. 

Let X X and Y Y, Fig. 83, be the principal axes of a section 
and let k^ and k„ be the radii of gyration about the two axes. 
With o as centre draw an ellipse, o x being equal to ky and o Y 




Fig. 83. — Momental Ellipse of Ellipse of Inertia. 



being equal to k,.. Then this ellipse is called the momental 
ellipse or ellipse of inertia. 

To obtain the radius of gyration k._ about a line z z passing 
through o at an angle ^ to x x, draw 2; z a tangent to the 
ellipse parallel to z z, and draw o Q perpendicular to it. 

Then o q^ = k. 
for I,^ = 2 a . p r2 

= 2 a (P S — S R)2 
= 2 a (P S — N t)2 

= 2 tt (1/ COS 6 — X sin O)"^ 

= -%a.x^ sin2 + 2ay^ cos^ ^ - 2 2 a; ?/ sin ^ cos ^ 

= sin2 e^a.x^ + cos2 O^ay^ -2sinecos01xy 



GEOMETRICAL PROPERTIES OF SECTIONS 171 

Now, 2 X y is the product moment, and as x x and Y Y are 
the principal axes, this is zero. 

.• . I,, - sin2 e^{a.x^) + cos2 e^(a 2/2) 
= I^^ sin^ $ + I,^ cos^ 
.-. Ak,^ == A h^ sin2 + Ak/ cos^ 
k.} = A;/ sin^ $ 4- h,^ cos^ 

and therefore from the properties of the ellipse o Q is equal to k,. 
A rather more convenient construction (see Fig. 84) is to 




draw a circle with radius k^ and draw o d at right angles 
to z z to meet the circle in d. Draw d e horizontally to 
meet the ellipse in e, then o e = A;-. If many values are 
not required, the ellipse need not be drawn at all, but instead 
draw a second circle with radius k,, ; then draw F e vertically 
to meet d e in e, thus fixing the point e. 

To find the principal axes in the case where there is no axis 
of symmetry, the procedure is as follows — 

(a) By graphical methods or by calculation first find the 
value of the product moment and the radii of gyration about 
any two axes through the centroid at right angles. 



172 THE STRENGTH OF MATERIALS 

Let the product moment be A p'^ and the radii of gyration 
k_r and k„. 

Then the angle of inclination B of the principal axes to k^ or 
k„ are given by the relation 

{b) By graphical methods or by calculation find the second 
moments of the given figure about lines x x and Y Y at right 
angles to each other and passing through the centroid and 
find it also about a third line z z at 45° to the other two. 

Then if is the inclination of the principal axes to x x and y y 

tan 2 ^ = L^+l^-^J^. 

CoxDiTiox THAT Peodijct Momext IS Zero. — It can be 
shown that the condition that the product moment about two 
lines is zero is that such lines form conjugate diameters of the 
moment al ellipse. 

A numerical example on the momental ellipse will be found 
on p. 242. 

Second Moments about any Two Lines through 
the Centroid at Right Angles. — A property of the second 
moments of a figure that is sometimes useful is that the sum 
of the second moments of an area, about two lines at right 
angles through the centroid, is equal to the sum of the second 
moments about any other pair of lines at right angles through 
the centroid. 

Second Moment or Moment of Inertia of Figure 
about an Axis perpendicular to its Plane. — The second 
moment or moment of inertia of an area about an axis o 
perpendicular to its plane is called the polar second moment 
or moment of inertia, and is equal to 2 a . p o^. 

Let any two axes x x and y y at right angles be drawn 
through o, and let perpendiculars p x, p m be drawn to these 
axes, Fig. 85. 



GEOMETRICAL PROPERTIES OF SECTIONS 173 

Then p o'-^ = p n^ 4- n o^ 

= PN^ + P M^ 
. • . 2 a . P O^ --= :^ ft . P N^ + ^ a . P M^ 

= l\x + Ivv 




Fig. 85. — Polar Moment of Inertia. 

Therefore wc have the following rule — 

The polar second moment, or moment of inertia, about an 
axis perpendicular to the plane of any area, is equal to the 




Fig. 80. 

sums of the second niojiicnls about any two linos at right 
angles, drawn through the axis in the plane of the area. 

The Determination of Gentroids, Moments of 
Inertia, and Radii of Gyration. — (a) Mathematical. — 
Consider the curve of a function y = V (x). 



174 



THE STRENGTH OF ^UTERIALS 



Then considering a strip of width d x parallel to the axis of 

X, Fig. 86 

Area of curve = / F {x) d x 

First moment of area about o y = / F (x) dx x x 

Second moment of area about o y = / F (x) dx x x^ 

Consider, for example, the parabola y- = 4:a x, and take the 
area between the curve and the axis of x, Fig. 87. 

Y 2 




Fig. 87. 



Area of curve = fy dx = / 2 a^ x^ 

B 

= 2 a^J'x^dx = \ 2a^- . 



dx 



2 .... 
„ x-^ 



a^ B- 



Now 2 «' b' = H 



Area of curve = ^ b H. Fig. 87. 



First moment about o Y = fx y dx ^ J 2ah x^-dx 

2ah / x^dx ^ 2 «K ; x^ 



4 2 

^ah B^ = ^ b^h 
o o 



-1 B H 6 

. • . distance of centroid from o Y == ^V—— : = c B 

o' B H O 



GEOMETRICAL PROPERTIES OF SECTIONS 175 
Second moment about o y = Cx^ y dx = f2 oh xr^ d x 



= 2 a^J x^ dx = 




2aKl x^ 
7 _ 




4 , : 23 

= ^ a^ B^ = ^, B^ H 


• ^2_tB^h 

|BH 


or k^ = Vf B. 


If the second moment is required about the base x z, we 


proceed as follows — 


T 2 3 
lov - ;^ b3 H 


I,, =. I,, -k.d^ 


232 9b2 

- -^ H B 3 • B H . 25 


I.\z ^ Ice + A. »]_ 


2,6 „ , 8 3 
= 7^^ -25^^ +75^^ 







^ fl50 - 126 + 561 „ 80 

\ 525 ^ ^^ 525 



H B'* -I ,.7^-^ h = H B' 

16 
105''^^ 



A list of values of second moments, etc., for common figures 
will be found on p. 185. 

It often happens in practice that the mathematical method 
is unworkable, in which case the following graphical methods 
are necessary. 

(6) Gbaphical. — First and Second Moment Curve 
Method. — (1) Centroid. — Suppose we have any area p r q s, Fig. 
88, and any two parallel lines x x and y y, at distance h apart. 

Draw a thin strip of the area parallel to x x and of thickness 
t and let its centre line be p Q. From one of the ends of this 
centre line, say Q, draw a perpendicular Q m to Y Y and from 
the other end draw p N perpendicular to x x. 

Join M N and let it cut p q in q^ and produce m q to cut 
x X in L. 



176 



THE STKENGTH OF MATERIALS 



Then the As p n q^, m n l are similar. 



p Qi _ NJ. _ P Q 

P N ~ M L ~ h 



PQi 






Multipljdng through by t we have 



p Qi X ^ = 



P Q X ^ X PN _ area of strip p Q x p n 



h 



h 



A^^^-P^ 4.- - £ ^ • First moment of strip about XX,,, 

Area oi portion p q^ of strip = , ^ (1) 




Fig. 88.— Graphical Determination of Centroid and Moment of 

Inertia, etc. 

Now divide the whole area up into strips and join up all 
the points corresponding to Q^, thus obtaining the First 
Moment Curve R Q^ s. 

Then the area to the left-hand side of the first moment 
curve Avill be the sum of the areas of portions of strips such as 
p Qj. Call this the First Moment Area (A^). Then we have 
Sum of first moments of strips about x x 



Ai = 



First moment of whole area 
h 



GEOMETRICAL PKOPERTIES OF SECTIONS 177 

. * . First moment of whole area = A^ ^ 

, . , „ , . 1 p First moment about x x 

or distance oi eentroid irom x x = '^ 

area oi figure 

= -i^ (2) 

Draw any vertical line f b to cut x x in r and y y in b, and 
through F draw any inclined line, on which set out f a equal on 
some scale to A, and f a^ equal to A^. Join a B and draw 
a^ c parallel to it, then the eentroid lies on a line through c 
parallel to x x or Y Y. 

^ C F F tti 

For — = — ± 
F B Fa 

C F _ Ai 

• ' • h ~ A 

A.h 
or c F = ~\~ 
A 

And this by relation (2) above gives the distance of the eentroid' 
from x X. 

(2) Second Moment. — ^If the second moment is required 
about the line x x draw q^ m^ perpendicular to Y Y and join 
Ml N, cutting p Q in Q2 and let m^ q^ produced cut x x in l^. 

Then the As p n q^, m^ n l^ are similar. 
• ^^2 ^ J^ Li ^ PQi 

* * P N Mj Li h 

P Qi X P N 
.-.PQ, =^^^ 

Multiply through by t, then we have 

P Ql X i X P N 



P Q2 X ^ = 



h 



T, , 1 ,, , , area of strip p q x p n 
±>ut we have seen that p Qi x ^ = f 



p Q2 X i 



area of strip p q x p n^ 



second moment of strip p Q about X N 
.*. Area of portion p Q.2 of strip = —7-2 . .(o) 

Now repeat this construction for each of the strips and join 
up all the points corresponding to Q2, thus obtaining the 
second moment curve R Q2 s. 

N 



178 THE STRENGTH OF MATERIALS 

Then the area to the left-hand side of the second moment 
curve will be the sum of the areas of portions of strips such as 
p Q2. Call this the second moment area (Ag). Then we have 
. _ Sum of second moments of strips about x x 

.-. Ix. = A2A2 (4) 

Some care is required in determining which area to read as 
A^ or Ag. It does not matter whether the verticals are drawn 
downward from p or from Q, but when the moments are re- 
quired about one of the lines, say x x, read, for the first 
moment area, the area on that side of the first moment curve 
from which the perpendiculars are drawn to x x, and in 
drawing the second moment curve draw from the first moment 
points, such as Qj, perpendiculars to the other line y y, again 
reading the area to the side from which the perpendiculars 
were drawn to x x. 

Now, on the line f a set out f a^^ equal to Ag on the same 
scale to which the other areas were drawn, and join a^ b, 
drawing ag d parallel to it. 

On D F describe a semicircle, and draw a line c e parallel 
to X X to meet it in e. 

Then c e will be equal to k, the radius of gyration about c c. 

Proof — 

F D _ F ^2 _ ^2 



Now 



F B 


F «! 




Ax 






F D 


li X Ag 




Jl X Ixx 


A .CF 


k/ 




A X CF 


CF 








h 






F D 


F E 










F E 


FC 










• • 


F D . F C 




F E^ 








.* . F D 


=r 


FE2 
CF 








.*. F E 


= 


kr 







GEOMETRICAL PROPERTIES OF SECTIONS 179 



Now F E^ = F C^ + C E^ 

. • . k,^ = c E^ + d_r^ 
. • . c e2 - yfc,2 _ ^2 

But we have already shown that 
k} = k/ - d/ 
. • . c E = A;,, 

Numerical Example. — Graphical Determination of Radius 
of Gyration of Rail Section about Centroid. 



x^./lrea of HalfSecfJoni. 
'■ = •ici-/8-Sti.ln I 

7 



Sum Curve o/-|-J.-\-L'?_r^ 
Halj Rail 4 T^i 1 V 




Fig. 89.— Rail Section. 

Fig. 89 shows the graphical determination of the radius of 
gyration about the centroid parallel to the base of a British 
Standard 85 lb. flat rail section. 

Since the section is symmetrical about a vertical centre line, 
the first and second moment curves need be drawn only for 
half the section, this simplifying the construction considerably. 
The lines x x and Y Y are taken as the horizontal lines, touching 
the section at top and bottom. 



180 



THE STRENGTH OF MATERIALS 



The areas A, Aj, Ag are next found b}^ planimeter or by sum- 
curve construction. (To avoid complication of figures, the 
sum curves for the first and second moment curves are 
omitted.) The first and second moment areas are to the left 
of the curves. 

When multij)lied by two, because only half the section was 
considered, we get 

A = 8*36 sq. ins. A^ = 4'02 sq. ins. Ag = 305 sq. ins. 

To the side of the figure a vertical r b is drawn between the 
X X and Y Y lines, and the points a, a^, a^ obtained as shown. 




X D C X 

Fig. 90. — ^IMoment of Inertia of Rectangle. 



Then by joining a b and drawing a^ c parallel, we get the 
point c, c F giving the distance of the centroid from x x ; and 
by joining a^ B and drawing a^ d parallel, we get the point d. 

On D F a semicircle is drawn, and c e is dra^\ii horizontal 
to meet the semicircle in e. 

Then c e = k, which on measuring will be found to be 
1-91 ins. 

This construction should be gone through as an exercise. 

Application of above Method to Case of Rectangle. — 
Let A B c D, Fig. 90, be a rectangle of base h and height h, and 
take the lines x x and y y through c d and b a respectively. 
Then the first moment curve will be the diagonal bed, Avhile 



GEOMETRICAL PROPERTIES OF SECTIONS 181 

the second moment curve will come a parabola b f d, so 
that 

A, ''^ 



A. 



2 
bh 



d.. = 






A 



2 bh 




Fig. 91. — Mohr's Construction for Moment of Inertia. 

.-.I., = I,, - Ad/ 

_bh^ _ bh.h^ _bh^ 
~ 3" 4 ~ T2 



Alternative Graphical Construction — Mohr's Method. 
— The following graphical method for obtaining the second 
moment about the centroid is in some cases more convenient 
in use than the one previously given. Divide the area, Fig. 91 , 



182 THE STRENGTH OF ^UTERIALS 

into a number of small strips of equal breadth, parallel to 
the direction about which moments are taken, and draw the 
centre line of each of said strips. Then if the strips are suffi- 
ciently small (we have only taken a few strips in the figure to 
avoid complication) the lengths of these centre lines represent 
the areas of the separate strips. Xow. on a vector line, to 
some scale, set out 0, 1, 1, 2, ... 6. 7 to represent the area of each 
strip, and take a pole p at distance = ^ total area 0, 7 from 
this vector line. Then anj'where across space draw and 
produce a line a k parallel to 0. p ; across space 1 draw a b 
parallel to p 1 ; across space 2, b c parallel to p 2, and so on 
until the point g is reached. Then draw the la?>t link g'h 
parallel to the last line p 7 to meet a 7i in h. 

Then the line c c through the centroid passes through h, 
and if a is the area of the shaded area, and A is the area of the 
figure, 

I- of figure = A >: a. 

Proof. — Consider one of the elemental areas, say 0, 1, and 
produce a b to meet the horizontal through k in b^. Then, by 
the law of the link and vector polygon construction, treating 
the areas of the elements as forces, 

1 

polar distance 





b'h 


= 


moment 


of first 


for 


'ce about 


c 


c 






= 


0, 


1 


X .r 




1 










^ ^to1 


:al 


area 








= 


0. 


1 


< X 













Area 


of . 


la 


b' 
0. 


h 
1 


_ 1 

2 
X X 


b' h X . 

■2x2 


r 









2 A 
second moment of element about c c 

, , 11^ second moment of fisure about c c 
. • . Area of shaded figure = a = r-^ 

?'. e. A a = second moment of ficfure about c c. • 



GEOMETRICAL PROPERTIES OF SECTIONS 183 

The proof that Ji determines the centroid is based upon the 
fact that in the hnk and vector polygon construction the 
meet of the first and last links determines the resultant, and 
in this case the centroid is where the resultant of the separate 
areas considered as forces act. 

* Equivalent Centroid and Second Moment of 
Heterogeneous Sections. — Suppose that the cross section 
of a beam is composed of two materials for which Young's 
modulus is not the same, and let Young's modulus for one 
material B be m times Young's modulus for the second 
material C. Then in the case of direct stress we have seen 
that the material B behaves as if it were replaced by m times 
its area of the material C. In the case of a beam the same 
relation holds, so that we may replace the material B by an 
area m times as wide, the width being taken parallel to the 
line about which moments are taken. 

Then if A is the area of material B, and A^ that of material 
C, the equivalent area of homogeneous material C is given by 

A2 = Ai + m A 

To obtain the distance d of the equivalent centroid from a 
line X X, take first moments of the separate areas about x x 
and let them be M and M^ respectively. 

Then equivalent first moment of the second material is 

M2 = Ml + m M 
Ml + mM 
Aj + w A 

To obtain the equivalent second moment about a line x x, 
take the separate second moments about x x and let them be 
I and Ii respectively, then the equivalent second moment of 
the second material is given by 

I2 = Ii + m I 

We shaU give numerical examples and further explanation 
of this when dealing with flitched beams and reinforced 
beams. 



184 



THE STRENGTH OF MATERIALS 



The above reasoning ma}' be shown graphicall}" as 
follows — 

Let A B c D (Fig. 92) represent any area which has em- 
bedded in it two bars x and y of different material. For 
considering the moments about any line such as D B shown 
dotted, make a strip e r of the same depth as x, and of area 
equal to {m — I) area of x and also a strip G h of area equal to 
{m — 1) area of Y. 

Then the equivalent first and second moments of the 




heterogeneous section about the given line will be the same as 
a homogeneous section of form aefbghcd. 

We take e r = (?7z — 1) area of x because the bar already 
occupies an area equal to its area, so that equivalent area 
of second material = [{m — 1) -i- I] area of x = m x area 
of X. 

Calculation of Moment of Inertia and Radii o! 
Gyration of Sections used in Constructional Work. — 
The moments of inertia of sections composed of sections of 
known moment of inertia are found by adding up the moments 
of the separate parts, or subtracting when the area consists of 
the difference of known areas. 



GEOMETRICAL PROPERTIES OF SECTIONS 185 





1 








CO 
















CO 




W 1 


+ 














N 


^0 <N 


I 


-O :00 




rr |(M 


I 


CO Ti< 


1 


1 




l-l 


r< ^ 


1 




CO I 


^O 1,-H 


1 




CO 


' 


' 












r< <M 






















roH 




















^ s. 






















i-H 






















+ - 






















li-H 












c3 




" «.., 


" «., 


" — 






"* 


'^ 


CO 






U 


h< iM 


f< C<l 


r«S ,«0 


+ S 


^ (N 


^ -* 


"■m'* 


hC) 


1 


1 1 


1-1 


^0 ^ 


.^0 '^ 


^1^ 


'O ,-H 


t. «=> 


^l« 


lO 


1 










<M 






t; 1 


00 t- 




1 >~* 










^ 








r-H 




' «t-l 






















i o 










CO 












4^ 










r« CO 












(3 










^ ^ 












a 






















o 








CO 1 










eo 




g 


M 




' 


3^ 


' 


1 


' 


' 




1 










1— 1 






















+ 




"* 1 




^ 






M 


m 1 


'^ CO 




CO 


1 


»o 1 


1 


^ ,-H 


1 












Si 


I-H 






















hO 














i 


o 




















5 


rOl-?^ 


1 


^0 i(M 


1 


1 


i 


1 


^0 

CO IQC 


^ lTt< 


-ti 


M 




















a 


^ 




















6 










































«<-( 










^^ "^^ 












o 


><) 








I— 1 

I-H 














.2 


a 


>^ ica 


r< Cs| 


r^ [jO 


t+ 


1 


-^ |C<1 


1 


^Se 


^12 


S 


o 








^ 


1 




1 


(r? iio 


Pm 


fH 






























»< Ico 




















"^ 












i Area 






2!^ 


+ 

r— 1 


s> 






(M |C0 


r-l |C0 










S^ 














ID 












s ^ 


^^ 


Figure. 


1 

O 
* 0) 




(D 


02 


Fh 




<D o 

o rt 


iD O 






PLH 




^ 








PM 


Fh 

Ph 

05 


o 


1— 1 


(N 


CO 


Tt( 


lO 


CO 


I> 


00 



186 THE STRENGTH OF MATERIALS 



Y 


2 

L— b ^ 


.© 














t 










X 

r 






X 





Fig. 93. — Properties of Common Figures. 
For the properties of British Standard Steel Sections, see 
Appendix. 



GEOMETRICAL PROPERTIES OF SECTIONS 187 

The following examples should make the method of calcula- 
tion clear for any such case. See Figs. 93 and 94. 





f 


1 






h 


c 

-1 


< 
1 


c 






V 




-at- 



O 







I 



— 2_ 



10' 



^^ 



3-5 




•© 



•475 



•4 75' 



X 



-iz 



•375 







5 



/2i' 



Y 



4i' 



^' 



10' 



X 



Y 




Fig. 94. — ^Moments of Inertia, etc. 



188 THE STRENGTH OF MATERIALS 

(1) Box or I Section. — These are geometrically equi- 
valent as far as the line c c is concerned, because if the box 
section be cut in half vertically and the two halves be turned 
back to back, we get the I section. 

Then I„, = b^'-ib-lh){h-2tf 

(2) Hollow Circular Section. 

When the thickness of metal is small and equal to t, this 
approximates to I„ = ^^^ 

(3) Channel Section (neglecting inclination of sides and 
rounded corners). — Consider the section shown in Fig. 94 (3). 

Area = A = 35 x 475 + 505 x 375 + 35 x 475 
= 5*219 sq. ins. 

To obtain distance d^ of centroid from x x take first moments 

about X X. Then 

A X d, 

An^ 3-5 , 505 x -375 x 375 , 3-5 x '475 x 35 

= 3*5 X -475 X ^ H ^ ^ 2 

= 2-910 + -363 + 2-910 = 6183 

• ■ ^" = 5T2Y9 ^ 1185 ins. 

Second moment about x x = I^^ 

_ -4 75 X 3- 53 5-05 X -3753 -4 75 x 35 ^ 
- ^ + 3 +3 

= 6-775 + -089 + 6-775 = 13-639 in. units. 

= 13-639 - (5-219 x 11852) 

= 13-639 - 7-323 = 6-316 in. units. 

/ I 
.-. ^ = ^/ = 1010 nis. 
\ A 

(4) Cast-iron Beam Section. 

Area = A = 2 x IJ + 7 x 1 + 6 x li = 19 sq. ins. 



GEOMETRICAL PROPERTIES OF SECTIONS 189 

Moments round base 

A d,. = 3 X 9-25 + 7 X 5 + 9 X -75 

= 27-75 + 35 + 6-75 = 69-5 in. units. 

. • . d^. = ^^- = 3-658 ins. 

- 6 X 3 - 6583 _ 5 X 2-1583 2 x 6-342^ _ 1 x 4-892^ 
. • . Ice - 3 3 + 3 3 

= 219-95 in. units. 

(5) Built-up Mild Steel Column Section. — Composed 
of two 10 X 3J X 28*21 channels and four 12 in. x \ in. 
plates. Required to find k^ and ICy. From the Table of Stand- 
ard Sections we obtain the following information concerning 
the Channel Sections — 

Area of each 8-296 sq. ins. 
I about centroid parallel to x x = 117-9 in. units 

1 ,, ,, ,, y Y = O 194: ,, ,, 

Distance of centroid from web = -933 in. 

.-. Total area of section= (4 x 12 x i) -f- (2 x 8-296)= 40-592 sq. ins. 

Moment of Inertia about X X. 

2 channels, 117-9 each = 235'8 

2 pairs of 12 x | in. plates about centroid = ~ ^ — = 2-0 

K X d^ for two pairs of plates = 2 x 12 x 5-5^ =726-1 



Total . . . . . . = 963-9 in. units 



7 / 963-9 . „. . 

•*'^'==V4ir592==^'^^''''- 

Moment of Inertia about Y Y. 

4 X - X 12^ 
4 plates 12 X J about centroid = ~~ := 288-0 

2 channels about centroid = 2x81 94 =16-4 

A X ^2 for each channel = 2 x 8-296 x 3-1832 = 168-5 
Total = 472-9 



4729 
^^ ^ V 40-592 ^ ^'^^ "^^* 



190 



THE STRENGTH OF ^UTERIALS 



(6) Built-up Beam Section. — Composed of two 14 in. x 
6 in. 46 lb. I beams and four 14 in, x f in. plates (Fig. 95). 
Required I^^. 

From the Standard Section Tables we obtain the following 
information concerning the I beams — 

Area of each = 1353 



■•.XX 5 > 5 > 



= 440-5 



Mean thickness of each flange = 698 in. 




a 



Fig. 95. 



I^^ OF WHOLE Section (not attow^no for Rivets). - 
Ivx of two I beams -= 2 x 440o = 881 



■' X 14 /5\ 
I of two pairs of plates about centroid ~ - — - — x ( - ) 

12 \4/ 



4-8 



A d^ for two pairs of plates = 4x 14x — x 7 62o- = 2035 



Total 



2020 8 



AixowANCE FOR RivETS (ucglcct I of each rivet -hole about 
its centroid). 

Area of each hole = (2 x -^- -f '698)4 = 1TU4 

o o 

Dist. of centroid from xx = 7276 
.-. I,, = 4 X 1-704 ■ 7-276- = 3608 
. ■ . Xett I,, = 2920-8 - 3608 = 2560 



GEOMETRICAL PROPERTIES OF SECTIONS 191 

(7) Built-up Sections — Approximate Method. — The 

moment of inertia of built-up sections can be found approxim- 
ately by adding the moment of inertia of the I beams or 
channels to A d? for the plates, d being taken as the distance 
from the centre of one set of plates to x x and the nett area 
of the plates being taken for A. 

Taking the section of the previous example, we then get 
I^^ as follows — 

I„ of two I beams = 2 x 440-5 = 881 
kd^ for plates = 4 x ^ (^14 - 2 x ^^ x 7-52 ^ 1875 

Total approximate I^^ = 2756 in. units 



I 



CHAPTER VII 

STRESSES IN BEAMS 

We have seen in a previous chapter how the bending 
moment and shearing force at different points along a beam, 
loaded in various manners, can be found ; our next problem 
is to find the relations between these quantities and the stresses 
occurring in the beam. 

We shall get a good preliminary idea of the stresses occurring 
in beams b}^ considering a model devised by Professor Perr3\ 
Suppose that a beam fixed at one en(J carries a weight, W 
(Fig. 96), at the other end, and that it is cut through at a 
certain section. Then the right-hand portion can be kept in 
equilibrium by attaching a rope to the top and passing over 
a pullej^ a weight W being attached to the other end of the 
rope, and by placing a block B at the lower portion of the 
section and a chain a at the upper portion. Then the pull in 
the rope overcomes the shearing force ; and the block b carries 
a compressive force c, and the chain a carries a tensile force T. 
Since these are the onh' horizontal forces, they must be equal 
and opposite, and thus form a couple. Then the moment of 
this couple nuist be equal and opposite to the couple, due to 
the loading, which we have called the bending moment. 

In the actual beam, owing to the deflection which takes 

place, the material on one side of the beam will be stretched, 

and the material on the other side will be compressed, so that 

at some point between the two sides the material will not be 

strained at all, and the axis in the section of the beam at which 

192 



STRESSES IN BEAMS 



193 



no strain occurs is called the neutral axis (N.A.). We see, 
therefore, that : The neutral axis is the line in the section 
of a beam along which no strain^ and therefore no stress, 
occurs. 

In an elevation of a beam there is also a line of no strain or 
stress, which may also be termed a neutral axis. These two 
axes are really the traces of a neutral surface. 

If we kno^ the manner in which the strain varies from the 
neutral axis to the outer sides of the beam, from a knowledge 
of the relation between stress and strain we can find the 
stresses at different points across the beam, remembering that 
the total compressive stress must be equal to the total tensile 





Te n 



Fig. 96. — Stresses in Beams. 

stress, and the moment of their couple must be equal to the 

bending moment. The moment of the couple due to the 

stresses is often called the moment of resistance. 

Assumptions in Ordinary Beam Theory. — We will first 

make the following assumptions with regard to the bending 

of beams, and from such assumptions we will deduce a relation 

between the maximum stresses, due to bending at any cross 

section and the bending moment — 
o 



194 THE STRENGTH OF IVIATERIALS 

(a) That for the material the stress is proportional to the 

strain, and that Young's modulus (E) is equal for 

tension and compression. 
(6) That a cross section of the beam which is plane before 

bending remains plane after bending, 
(c) That the original radius of curvature of the beam is very 

great compared with the cross-sectional dimensions of 

the beam. 

We will also for the present restrict our investigation to the 
case of simple bending, ^. e. that in which the following con- 
ditions hold — 

(1) There is no resultant thrust or pull across the cross 

section of the beam. 

(2) The section of the beam is symmetrical about an axis 

through the centroid of the cross section parallel to 
the plane in which bending occurs. 

To get a clear idea of the stresses in beams it is absolute^ 
necessary to have a clear idea of the assumptions involved in 
formulating any particular theory, and of the effect of such 
assumptions on the results. 

Let A B, Fig. 97, represent the cross section of a beam which 
has been bent (the amount of bending having been exagger- 
ated). Before bending, the line a b had the position of a^ b^ 
so that B Bi represents the maximum tensile strain, and a a^ 
the maximum compression strain. From our assumption (6), 
called Bernoulli's assumption, a^ b^ and a b are both straight 
lines. The neutral axis then passes through c, the point of 
no strain, and it follows from the above assumptions that the 
strains are proportional to the distances from the N.A. From 
assumption (a) it foUows that the diagram of intensity of 
stress is also a sloping straight line, Ag Bg, the portions Bg c 
and Cg A being continuous, because Young's modulus is equal 
in tension and compression. 

It is clear that the maximum stresses in compression and 



STRESSES m BEAMS 



195 



tension occur at the points A and b, and let these be /,. and ft 
respectively, d, and dt being the distance A c and B c . 

Position of Neutral Axis. — Now consider an element of 
area a at a point p at distance p n from the N.A. 

Then the stress at the points p is equal to Pj Pg 




Tension 



Cross -section Diagram of 

Intensity of Stress 

Fig. 97. — Stresses in Beams. 



But 



Pj^ C AC d, 

fc 
Pi P2 = J X Pi C 

= -^^ X P N 

d. 



. • . Stress carried by the element — a xH x P N 
.' . Total stress carried by section above N.A. = ]Sax^XPN 



fr 



= ■^ 2 a X p N 
dr 



= , X first moment of area above N.A. about N.A. 
dr 

Similarly if an element of area at a point p^ be considered, 

we see that 



I 



196 THE STRENGTH OF IVIATERIALS 

Total stress carried by section below N.A. 

= 3^ X first moment of area below N.A. about N.A. 

But we have seen that the total tension T must be equal to 
the total compression C, and it follows from assumptions 
(a) (h) that 

d,. d, 

.'. we see that the first moment of the areas above and below 
the N.A. about the N.A. are equal and opposite in sign. 
Therefore, the total first moment of the whole area about the 
N.A. is zero. But we have seen that the first moment of an 
area is zero about a line through the centroid. 

Therefore, i7i simple bending with the given assumptions, the 
neutral axis passes through the centroid. 

The Moment of Resistance (M.R.) — We have proved 
that the stress carried on an element a of area about a point 

p is equal to a x j.' x p n 

The moment of this stress about the N.A. 

= stress X P N 

= a X ,' X pn2 
d, 

. ' . Total moment of all the stresses over the cross section 

= 2 a . ^' X P n2 
a, 

' = ^' :S (a X P n2) 

^ J (second moment of ^^hole area about the N.A.) 
a,. 

d. 
But the total moment of all the stresses is the moment of 
the couple w^hich we have called the moment of resistance. 

.*. we see that M.R. = , or ' 

d, a, 



I 



STRESSES IN BEAMS 197 

The moment of resistance must, as has already been shown, 
be equal to the bending moment, which we will call M. 

.-. M ^y^or V (1) 

It will be seen that I, d, and dt depend merely on the shape 

II 

of the cross section, and -,- and -,- are called the compression 

U/Q (hi, 

modulus and tension modulus respectively of the section, and 
are written Z, and Z<. 

Thus our relation becomes 

M = /,Z, =/,Z, (2) 

In practice we usually want to know /<, and ft which give the 
maximum stresses across the section, and so we will write the 
result as 

/,. = ! (3) 

M 

f'-% (*^ 

In the case where the section is symmetrical about the N.A., 
d^ is equal to dt, so that Z^ and Z^ are equal. In this case, 
therefore, /, = ft, and we may write the relation as 

' Z 
For values of section moduli for British Standard Beam 
Sections, see Appendix. 

Unit Section Modulus ; Beam Factor. — If instead of 

taking the section modulus as Z we took -v and called it the 

" unit section modulus," the quantity would be rather more 
useful and probably more easy to conceive to those who find 
difficulty in purely mathematical conceptions. 

We should then have 

Z I A;^ 

Unit compression section modulus = z, = -^ "= a ^ "^ ~d~ 

„ tension „ ,, = Zt = -j 

z, and Zi then become lengths 



1*\^ 



rnK v^TKKNcrrn ov MVTK)n\i.s 



oi|iK-\tious (o^ ;\iul {4) ^ivi^ 



A 



/,.A 



'A 

M 
Z, 



«, ivud ;< oau bo found giaphioally bv sotting ont c p. 
Fig. l>7<i. horizontally to ivpix^sont k , tho radius of gyration 




Fu.. i>7i». 

i>f tho soot ion abovo tho >*.A.. joining u i\ \ v and drawing linos 
OK, DF ivsptvtivoly at riglit anglos to thoni. Thon i' k, c f 
are ;:, and ;. ivsjHvtivoly. 

Take, for oxaniplo, tho As r k i\ nc o: thoy aro similar. 



6Vum factor. — Tho quantity "J = v r ^*^^^ '^ synunotrioal 



STR-KSSKS IN liKAMS ID!) 

,. Z rniM. ,. , „ i • I i- 1 

K(H!ti(tri, ()»• . , , lor- n, ])<",ih\ <»l ;i.Hyirimcl,ncal Hcotioii wIi(>h(5 
A a 

Hafo bfMiding Hl/fOHHCiH an^ ilio HarrKJ in cotuprc-HHiou ;iikI (-criHioii, 

i.M Ji. (iicjiHiirfi of Mic vJiJiK! of I-Im! Hf5(!l,ioji n,H Ji f)ca,fri u.hd lun.y 

!»(«, cMlicd Mi(5 />(,(wi Idcior. 'i\il<(i, for inHJyarujf^ l/hc- If/' / (^ 

Hliuidard I Ixwirri fA()|M-ii(lix). Z 8:*>!) ;i,nd A I7'{. 

.'. boarn la(;l>or .^ •'J23 

For IIk^ W / H^'Hiandard boairi Z ()()(), A 2()-2. 

.". beam factor ,., ' „,, ,, ".'{42 
10 X 20-2 

.'. tho 10'' X 8'' in a mom rusoMomicjal Hnoficm. 
Ar(5(5ian^ular HccMoii wonid ^dvc- u, beam facjtor ^ ^ \ "1^7. 

N(JMici{.K!Ai, i*]xAiviiMJO,s. - 'rii'5 following fi umc/fical ('/xam[)l(5H 
will ma,kc- il, cJcai- liow llio HlroHHOH in bcarriH loa,dc,d in ^dv<ri 
inanncrH nan })0 found, Jind liow a HaT(j lo;id can Im-, found for 
a hcarri of ii'w/cw Hf)a,fi a,n(J wjcjtion. 

(I) ^rihc. five se.ciiona a, h, c, d, c, Fuj. 08, have, each an area o/ 
4 wf. i'nfi. Find their relative Hl/ren<jlJhn an heamM for Ike H(i,w,e 
H'jxi.n, if they a,re of the nawe m.aleriaL 

Wo liavfi H('.(!n \,\\\\\, M /Z. Now if a,ll Uio bfiamH arc, 
loadc-d in \]\v, same way, M will b(; y)roporiional to the- load ibey 
(!an (larry, arid an / \h 1/ho Hamc, for cacb, w(5 hc.c- <-bat Ififrir 
rvlalivc Hlronj^UiH a,H b(;amH dcpc/nd on \\\c'w vaJuc-H of Mir; 
niorlnhis of each HC-cfion. For tabic of hcco/kJ mom(5ritH, 
H('(! |). I 85. 



Sert,i(fn a. 

h /,;■•' 2 / 2'' 
12 ''' 12 



i. 

Z 

.-. Z 



I I 

d I 

2 / 2'» 2/8 
12 y I 12 

r'».'> in. unilH. 



200 



THE STRENGTH OF IVUTERIALS 




Example- I 




Examble 5 



6 M. on each qirJer 

Fig. 98. — Examples of Beams. 



STRESSES IN BEAMS 201 

Section b. This is composed of two triangles. 

•.1 = 2 X ^^,h in this case being the height of the triangle. 

_ 2 X 2-828 X 1-4143 
' 12 

d = 1-414 

7 _ 2 X 2-828 X 1-4JL43 _ 2;82^ 
' 1-414x12 ~ 3 

= -943 in. units. 

Section c. 

T _^ _^ x_2-26^ 

64 ~ 64 " 
d = 113 



64 X 113 




= 1*13 in. units. 




Section d. 




2 X 43 2 X 
^ ~ 12 


•8 X 2-53 
12 


= 10-67 - 208 = 


= 8-59 


d = r 




. ^ 8-59 




= 4-29 in. units. 





Section e. This is composed of three rectangles. 

T = •'75 X 23 2-5 X -43 -75 X 23 
12 "^ 12 "^ 12 
= -5 + -013 + -5 
= 1013 
d = V 
- . Z = r013 in. units. 



We see, therefore, that the order of the sections, from 
strongest to weakest, is d, a, c, e,b. 



t 



202 THE STRENGTH OF MATERIALS 

We may take it, as a rule, that the strongest beam for a 
given area of cross section is that which has a depth as great 
as is practicalljr possible, and which has as much as possible 
of the metal at the outer portions of the beam. 

(2) A girder of 20 ft. span carries a uniformly distributed 
load of 10 tons, and a central load of 4 tons. Find a suitable 
British standard beam section for the girder if the maximum 
stress is to be 7 tons per sq. in. 

Its maximum B.M. due to the uniform load will be equal to 

WZ 

-g (see Eigs. 59, 59a, Cases 2 and 3) 

10 X 20 X 12 . , 
= X m. tons 

= 300 in. tons. 

The maximum B.M. due to the central load = -— ^ 

4 

^ 4 X 20 X 12 

~ 4 

= 240 in. tons. 

These both occur at the same point, so that the maximum 
B.M. due to both loads = 540 in. tons. 

Now M - / Z 

2. e. 540 = 7 Z 

.-. Z -= = 77"14 in. units. 

On referring to the table of standard sections (Appendix), 
we see that the section having the nearest modulus to this is 
a 14 X 6 X 57 lb. section for which Z = 76' 12, and we will 
adopt this section as being suihciently strong. 

(3) A tank which weighs J ton and measures 10' x 6' x 3' 
is filled with water, and carried on three girders placed length- 
wise, so that each girder takes an equal weight. If the girders 
are &' x 3'' x 12 lb. Standard Beams find the maximum stress 
in each. {A.M.I.C.E. Altered slightly.) 



STRESSES IN BEAMS 



203 



Weight of water in tank = 



10 X 6 X 3 X 62-5 



tons 



2240 
= 5-02 tons. 
. • . Total weight carried by girders = 5*02 + '5 = 5*52 tons 

i,r ' T^i^r u • ^ ^'52 10 X 12 

.'. Maximum B.M. on each girder = — ^- x 

o o 

= 27-6 in. tons. 

Z for a 6" x 3" x 12 lb. beam is 6*736 in. units 

i 27-6 ... 
. • . / = 7r--o^ =41 tons per sq. m. 
' 6-736 ^ ^ mi 



Compression 




Ih-nsion 
Cost Iron £>earn 



fliichea E>Qam . 



Fig. 99. 



(4) A cast-iron beam is the shape of an inverted T, 9 ins. deep 
over all, width of flange 6 ins., thickness of web and flange 1 in. 
If its length is 12 ft. find what weight at the centre will cause a 
tensile stress of 1 ton per sq. in. in the flange. What would the 
maximum compressive stress then be ? {A.M.I.C.E.) 

First find the centroid and second moment of the section. 
(See Fig. 99.) 

Area of section = A = 9 x 1 + 5 x 1 = 14 sq. in. 

9 1 

1st Moment about base = A fZ = (9 x 1) x ^ + 2 (2 J x 1) x ^ 

- 40-5 + 2-5 = 43 

•'• ^ = 1^ = 3-07 ins. 
14 



204 THE STRENGTH OF MATERIALS 



2nd Moment about base 


1 X 93 2 X 21 X 13 

~ ' ~ 3 + 3 




= 243 + 1-67 = 244-67 


. • . 2nd Moment about 


parallel line through centroid 


= 1, 


= 1, - A (P 




= 244-67 - 14 X 3-072 




= 244-67 - 13207 




= 112-6 in. units. 


.-.z. 


112-6 112-6 
9 - 3-07 5-93 




= 18-99 in. units. 



Z, == -^— r = 36-67 in. units. 

. • . Safe B.M. in tension = /, x Z^ 

= 36-67 in. tons. 



Neglecting weight of beam itself, if central load is W, the 
4 



maximum B.M. is 



. • . Maximum B.M. = ^ ^ = W x 1 22^1 2 _ gg ^ j^. tons. 

4 4 

^^j 36-67 1-02 tons. 
' • ^' 36 

f X d 1 X 593 
The compression stress —-^-^ — ' = — qTat — ~ -^"^^ ^^^^ P^^ ®^- ^^- 

(5) A pitched beam consists of two timbers, each 9 ins. thick and 
16 ins. deep, and a steel plate placed symmetrically between them, 
the steel plate being 8 ins. deep and | in. thick. If E for timber is 
1,500,000 lbs. per sq. in. and for steel 30,000,000 lbs. sq. per in., 
find the maximum tensile stress in the steel plate when the maximum 
tensile stress in the timber is 1000 lbs. per sq. in. 

Determine also for the same intensity of stress in the timber the 
percentage increase of load the flitched beam will carry as com- 
pared with the two timbers when not reinforced with the steel 
plate. {B.Sc. Lond.) 

Using the notation given on p. 183, we see that 

_ 30,000,000 _ 
™- 1,500,000 -^^ (seei-ig. JJ). 



STRESSES IN BEAMS 205 

.*. The steel plate is equivalent to a timber 20 times as 
wide, i. e. a timber 15 x 8 ins. 

. • . For the equivalent section of timber for the whole 

flitched beam Ig 

_ 2 X 9 X 163 (15 - I) 83 
~ 12 "^ " 12 

= 6,144 + 608 
= 6,752 in. units. . 

For the timber beam not reinforced I = 6,144. 

When the stress in the timber at the outside of the section 

is 1000 lbs. per sq. in,, that 4 ins. below the N.A., i. e. at the 

maximum depth of the equivalent timber plate, will be 

4 

■^ X 1000 = 500 lbs. per sq. m. 

o 

But steel carries 20 times the stress in the timber for the 
strain. 

.'. Stress in steel = 20 x 500 = 10,000 lbs. per sq. in. 
For the flitch beam the equivalent modulus is 
6752 



8 



844 in. units. 



S4.4. V 1000 
.-. Safe B.M. in ft. lbs. = »^^ J^^ = 70^333 

6 144 
For the plain timber beam Z = ~~ — = 768 in. units 

o 

.-.Safe B.M. in ft. lbs. = ^^^ ^J^^^^ = 64,000 
.'. Increased B.M. carried by flitched beam = 6,333 

.-. % increase = ^^000 ^ ^^^ "" ^^ '^ "/^ 

We shall have further numerical examples on the stresses in 
beams at various points in the book. 
Approximate Value of Modulus of I Sections. — In 

practice girders are usually made of I section, because the 
most economical section is that in which as much as possible 
of the metal is placed in the edges or flanges. In this case an 
approximate formula for the modulus of the section can be 
found as follows : Let d (Fig. 100) be the distance between the 



I 



206 



THE STRENGTH OF MATERIALS 



centre of flanges of the section, the thickness of the flanges 
being t. Then if b is the breadth of the flanges, and t-^ the 
thickness of the web, we have 



I = 



B (D + tf (B - y (D - tf 



(1) 



.-. 12 I = B(D3 + 3D2^ + 3D^2_^^3)_(B_y (d3_3d2^_^3p^2_^3) 
= B (6 J)H + ^3) _!_ ^^ (j)3 _ 3 J) 2^ + 3 J) ^2 _ ^3) 

Now if t is small compared with d, f^ and t^ are negligible, 



Combression Fl^nac 



^ 



t 



h — ^B 



.-^ 



Web 



D 



3 






FUnc?C , 



ension nanci 
FiQ. 100. 






Npw Z = 



I _ 21 

2 



21 



(3) 



since i is small compared with d. 



1 — nearly 



STRESSES IN BEAMS 207 



...z=.t{-'+'.-('-l')}('-3 



= ^[6^t-^^''+hi>-3tt,-tt, + ^'^f} ..(4) 

= |6 B if + ^1 (d — ^) I to a first approximation, 

neglecting all remaining terms containing t^ oi tti. 
Now B X ^ = area of one flange = A 
and ^1 (d — ^) = area of the web = a 

=K^+S ('> 

Therefore we get the following rule : The modulus of an I 
section beam is approximately equal to the depth between the 
centres of the flanges multiplied by the area of one flange plus 
one-sixth of the area of the web. 

Discrepancies between Theoretical and Actual 
Strengths of Beams. — Many practical men have expressed 
considerable surprise that in testing beams the actual and theo- 
retical breaking strengths do not agree. A number of beams 
are tested, and a tension test is also made from the same 
material, and it is found that the load which, on the ordinary 
bending theory should cause the breaking stress in the beam, 
does not cause fracture, the amount of additional load depending 
on the shape of the cross section. This was the origin of the 
old " beam paradox," it being thought that the material must 
be stronger in bending than in tension. In fact, for cast- 
iron beams, an old erroneous theory which, for a rectangular 

beam, made M = — . — instead of - — ^ agrees con- 
siderably better with the breaking test than the modern 
theory. 

Now this discrepancy in the case of ductile metals is due to 
the fact that the ordinary bending theory is not applicable to 
breaking stresses, and no one who appreciated the value of 



I 



208 THE STRENGTH OF MATERIALS 

the assumptions made in obtaining such theory would expect 
the theoretical and actual breaking strengths to agree. This 
is because the stress is not proportional to strain after the 
elastic limit is reached. 

Some experimenters who have measured the deflections of 
beams have stated that for mild steel the stresses at the elastic 
limit do not agree, but that is due to a confusion between the 
elastic limit and the yield point, and to the fact that the 
deflections were not measured with sufficient accuracy. In 
Chap. I we saw that for a tension test of mild steel the 
elastic limit and yield point were quite close to each other; 
but in bending this is not the case, the yield point occurring at 
a considerably later point than the elastic limit. Considerable 
error, therefore, arises if the yield point in bending be taken 
instead of the elastic limit. If the latter be carefully measured 
it will be found that the stresses in tension and bending at 
the elastic limit agree very closely. This point is proved, 
incidentally, in the Andrews-Pearson paper on Stresses in Crane 
Hooks, referred to in Chap. XIX. The reason for the yield 
point coming some distance after the elastic limit in bending is 
that only the material at the extreme edges has been stressed 
up to the yield point, and the whole section will not yield until 
the material nearer the centre has become stressed up to the 
yield point. 

We see, therefore, that there is no discrepancy between theory 
and tests so long as the conditions laid down in formulating 
the theory are fulfilled. If those conditions do not hold 
beyond a certain point, then, after that point, we must get a 
new theory if we wish to calculate the stresses. 

These so-called discrepancies between theoretical and actual 
strengths of beams point to the desirability of choosing the 
working stresses for ductile metals in terms of the stress at the 
elastic limit, and not of the breaking stress — as we pointed out 
in Chap. III. — because if the working stress in a beam is, 
say, one-half of the stress at the elastic limit in tension, then 
twice the load on the beam will cause the elastic limit in the 



STRESSES IN BEAMS 



209 



beam ; if, however, the workmg stress be taken as one-fourth 
of the breaking stress in tension, four times the load will not 
cause failure, the exact load to do this being more, and 
depending on the shape of the section. 

Cast-Iron Beams. — The discrepancy in the case of cast- 
iron beams is due to the fact that the stress -strain diagram is 
not a straight line except for very low stresses and that for 
given strain the stress is appreciably less in tension than in 



ConobressioK7. 




Tension. 



Fig. 101. 



compression. The result of this is that the stress diagram 
becomes more like that shown in Fig. 101 ; the neutral axis 
becomes raised above the centroid and the diagram is curved. 
In this figure the actual stress is shown about one-half of the 
calculated stress. This effect will be most marked in sections 
such as rounds or diagonal squares with a large amount of 
metal in the web; it is least in I sections with the com- 
pression flange smaller than the tensile, i. e. the discrepancy 
between theory and practice is least in a well-designed section. 
* Diagonal Square Sections. — It is an interesting fact 
that we can obtain as follows the apparently paradoxical 



I 



210 



THE STRENGTH OF MATERIALS 



result that by cutting away part of a beam of diagonal square 
section we increase it strength. 

Referring to Fig. 101a and the table on p. 185 



1^ , of whole section = 



(V 2)4. 12 



_ ^ 

~48 

.'. Z of whole section = .o -=~ o = o^ 

48 2 24 




A 



Fig. 101a. 



2d^ 



1^ ^ of two triangles removed = 2 x ^^ (about their own 
centroids) + 2 . cZ^ (^ ^ " V)' ^*° ^^^^^ ^^ ^'^^'^ 

.*. 1 01 remammg section — aq ~ q ~ '^d'^yn — o ) 



2^2 



48 t 
Z of remaining section 



I6d^ 



D2.48/, 4:dY] 



D^ 



f 



48 (D -2(Z) [^ 3d4 



16 (Z* 24 cZ2 



D^ 



4rf\2) 

3d;/ 



Z of remaining section 
Z of original section 
D f _ 16^* _24cZ2 
B-2d\ 3d* d2 



1 - 



id\^] 
3d) f 



STRESSES IN BEAMS 211 



Now let - = X 

D 



ai..,{'-^*-"-('-r 

,, -^^^ {1 - 24 a:2 + 64 a:^ - 48 x^} 
( 1 — 2 cc) ' 



= 1 + 2 :r - 20 a:2 + 24 ^3 

= (1 - 2a;)(l + 4 a; - 12 x'^) 

= (1 - 2 a;)2 (1 + 6 x) 

d X 
This is a maximum when -y^ = 

ax 

i. e. 2 - 40 a: + 72 ^2 = 

^. e. 1 - 20 a; + 36 ^2 = 

i.e. (1 - 18 a;) (1 -2 x) = 

From this x = ^ or J and the maximum is for x = ^s 
because clearly x = ^ reduces the section to zero. From this 
we see that the strongest section is obtained by removing 
one-eighteenth of the depth from the top and bottom, i. e. one- 
ninth of the depth in all. 

When ^ = To 

8\2 4 256 , .^„ , 

Therefore the section f of the depth of the original section 
appears to have a strength 1'053 times as much, i.e. about 
5*3 % increase. The additional strength is, however, only 
apparent, because when failure starts at the edges we arrive 
at the stronger section. 

We do not know of any accurate tests that have been made 
to find to what extent this result holds in the actual beam. 
In a cast-iron beam of the original section under test a small 
crack will start on the tension side which will have the effect 
of cutting off one edge only. A complete study of this 
problem on a modified theory for cast-iron beams will be 
found in a paper by Mr. Clark in Proc. Inst. C.E. (1901-2). 



212 



THE STRENGTH OE MATERIALS 



* Influence of Shearing Force on Stresses in Beams. 
— It must be remembered, that up to the present we have 
considered only the tensile and compression stresses due to the 
bending moment. Besides these stresses there are tangential 
stresses due to the shearing force. The resultant stress at any 
internal point of the beam is the resultant or principal stress 



/^ 


ComJDression 




i 


' \ / 




i 


^ / 


1 


i 




/ 


7 






i 


\ 


/ : 


i „ ... 


^^ 


/ 




"^^-- 



Ihnsion ^ 
S.M. and Shear StrGSse'S 'RGSultanf' 

Fig. 102. — Principal Stresses in Beams. 

Oowbressior? 




Tension ^ 



Fig. 102 a. 



of the tangential and direct stresses, which resultant is fo.und 
as sho\\Ti in Chap. I. We shall deal in a subsequent chapter 
with the distribution of the shearing stresses across the 
section of the beam, but for the present we will assume that 
the shear stress is a maximum at the centroid and diminishes 
to zero at the extremities. Fig. 102 shows diagrammatically 
the shear and direct stresses across the cross section of the 
beam and also the resultant stresses which, as will be seen, 



STRESSES IN BEAMS 



213 



are parallel to the centre line of the beam at the extremities 
and are perpendicular to it at the centroid. 

If the principal stresses at various depths be found for a 
number of cross sections at various points along the span., and 
the directions of principal stress be joined up by a curve, we 
get a number of lines showing the manner in which the direc- 
tions of principal stresses vary from one point to another. 
Such curves will be found in Rankine's Applied Mechanics, and 
are of the form shown in Fig. 102a. 

In practice it will be found that, except for very short beams 




Strain Diaaram 




carrying heavy loads, the maximum tensile or compressive 
stress due to bending moment is usuaUy greater than the 
maximum shear stress, so that the consideration of stress due 
to bending moment is, as a rule, considerably more important 
than that of the shear stresses. 

* Moment of Resistance in General Case. — To follow 
the correct theory of beams it is not necessary to make any of 
the assumptions previously given, and we wiU now find the 
moment of resistance in the most general case. To investigate 
this, we must suppose that we know by experimental or other 
means the shape after distortion which is taken up by a cross 



I 



214 THE STRENGTH OF MATERIALS 

section of the beam which was originally plane. We must also 
know the relation between stress and strain for the material of 
which the beam is composed. 

Let A B, Fig. 103, represent the elevation of a cross section 
of a beam which after bending is strained to the shape d c e. 
Then from the stress-strain curve and from the shape of the 
cross section draw a curve of stress d' c e'. This is obtained as 
foUows : let a 6 be any ordinate of the strain diagram ; then 
from the stress-strain curve find the stress corresponding to 
this strain, and multiply the stress by the breadth of the beam 
at the given point, and plot this equal to a' h' to some con- 
venient scale ; joining up points such as h' we get the stress 
diagram. 

Now let the area of the stress diagrams be Q and T and their 
centroids G^ and G2. Then, of course, in simple bending Q and 
T will be equal, and if q is the perpendicular distance between 
the centroids, the moment of resistance will be equal to T x g 
or Q X g. 

If the reader fully follows this general method with regard 
to the stresses in beams, he should not have the difficulty 
commonly experienced in following the more particular 
theories. 



CHAPTER VIII 

STRESSES IN BEAMS — (continued) 



* 



REINFORCED CONCRETE BEAMS 



There are many formulae for the strength of reinforced 
concrete beams, such formulae being deduced from certain 
assumptions with reference to the distribution of stress in the 
bent beam. 

We will consider three methods of calculating the stresses in 
reinforced concrete beams, working in each case the case of a 
rectangular section, this being most common, and in all three 
we will make the following assumptions — 

(1) That a section of the beam which is plane before bending 
remains plane after bending. (Bernoulli's assumption (see 
p. 194).) 

(2) That the beam is subjected to pure bending, i. e. that 
the total compressive stress is equal to the total tensile 
stress. 

Standard Notation. — Throughout the treatment we will 
adopt the following notation (see Fig. 104). 

Young's modulus for steel or other metal E, 
Young's modulus for concrete E^. 

t = Tensile stress per sq. in. in reinforcement. 
t, = ,, ,, ,, concrete, 

c = Compressive stress per sq. in. in concrete. 
At = Area of cross section of reinforcement. 
Ac = ,, ,, „ concrete. 

b = Breadth of beam. 

215 



210 



THE STRENGTH OF MATERIALS 



df = Total depth of beam. 

d = Depth of beam to centre of reinforcement. 
n = Depth from compressive edge to neutral axis (N.A.). 
(^d — n) — ,, ,, centre of reinforcement to neutral axis. 

n-i = Ratio y. 

r = Proportional area of reinforcement to area above 

XXt, 



it 



bd 

1, = Equivalent moment of Inertia of section. 





-* b >- 


..1 


>\ 


n 


V 


fu 


■ 




^ 


L_.0 ^__ 




'f 



79 



Fig. 104. — Notation for Reinforced Concrete Beams. 

First Method — Ordinary Bending Theory. — The first 
method which we will consider is one which is not much used in 
practice because it gives safe loads which are lower than tests 
show to be necessary. It is, however, the general method 
applicable to beams formed of two elastic materials, and serves 
as a useful and instructive introduction to the subject. 

According to this method, we assume that the reinforced 
beam behaves exactly as an ordinary homogeneous beam with 
the reinforcement replaced by a narrow strip m times the area 
of the reinforcement, and at constant distance from the N.A. 

We showed how to find the centroid, moment of inertia, and 
radius of gyration of such an equivalent homogeneous section 
on p. 183. 

In the general case, let'7^^(^ig. 105) be the distance to the 



STRESSES IN BEAMS 



217 



neutral axis (equivalent centroid), and I^. the equivalent mo- 
ment of inertia about the centroid. 

Then*..=M(^;:=^) (1) 



c = 



t = 



Mn 



mM{d — n) 



(2) 
(3) 



where M is the bending moment. 



d 



N 



t 



n. 



d 



W 



Fig. 105. — Reinforced Concrete Beams. ■ Method 1. 

In the case of the rectangular beam we then get the following 
results — 

Equivalent area of section ^ h dt + {m — \) A., (4) 

As explained on p. 184, it is (m — 1) A^, because when we 
take away the reinforcement and replace it by m times its 
area of concrete, we have first to fill up the hole in which the 
reinforcement was, and this takes once A^, so that remaining 
additional area = (w — 1) A^. 

Take moments round the top, then we have 



n [h dt -^ {m — \) A, 



1 = -f '^ + (m- l)A,d 

''2'- +(m- 1)A,^ 
n = ,— , 



bd, + {m - I) A 
This fixes the position of the neutral axis. 



(5) 



218 THE STRENGTH OF JVIATERIALS 

Taking second moments about the neutral axis, we have 
I = ^f + ^-i^^- + (m - 1) A, (d - «r- (6) 

In this formula we neglect the second moment of reinforce- 
ment about its own axis. 

Numerical Exa]viple. — Take the case of a beam 6 ins. wide 
and 12 ins. deep, the centre of the reinforcement being 2 ins. from 
the bottom and the area of reinforcement = 1*44 {see Fig. 106). 




Fig. 106. 



Taking m = 15, we get 

_ 6 X 144 + 2 X 14 X 1-44 X 10 
^ ~ 2(72 + 14 X 1-44) 

= 6-87 ins. 
,'. {d-n) =^12 - 6-87 = 5-13 

I = Lx (6:87)3 6_x (5-13)3 ^.^^ ^ 3.^3, 

3 3 

= 626 + 270 + 197 = 1093 nearly. 

. • . Taking a safe stress of 100 lbs. per sq. in. in tension for the 
concrete, 

Safe B.M. = ^^^^{^f^ ^ ^ '^^ ^*- ^^'• 



100 X 6-87 



Then t, = comp. stress in concrete — ^ ,„ 

15 X 100 X 3-13 



Then t = Tensile stress in steel = 



513 



= 134 lb./in.2 
- = 915 lb./in.2 



STRESSES IN BEAMS 219 

It will be seen that we have taken t^ — 100, which is higher 
than usually allowed for concrete in tension ; but if the con- 
crete cracked the steel would still hold, and so we are justified 
in using a higher stress. 

The above example shows that on this method of calculation 
the beam is not very economical, as the steel is very little 
stressed and the concrete has only a small stress in compression. 

For this reason it is usual in practice to neglect the tensile 
stresses in the concrete, that is to say that it does not matter if 
the concrete does crack. Practice shows that such cracks, if 
present, do not matter so long as the adhesion between steel 
and concrete is good, and the tensile stress in the steel and the 
compressive stress in the concrete are within safe limits. 

We should like in this connection to point out that to neglect 
the tensile stresses in the concrete does not, as some writers 
state, increase the factor of safety. We shall see later that 
neglecting such stresses we get a much larger safe B.M. on the 
beam, and thus reduce the factor of safety. 

Strength of Same Beam not Reinforced. — To serve as 
a useful comparison we will find the strength of a 12''' x Q'^ 
concrete beam without reinforcement. 

M X 6 M 



If not reinforced, t = 



6 X 123 144 



12 

In this case we must take safe t^ = 50 Ib./in.^ 

. • . Safe B.M. = ^"^ ^^^^^ = 600 ft. lbs. 

Therefore, calculating by our first method, the reinforced 
beam is roughly three times as strong. It would cost roughly 
twice as much, so that we see there is 50 % saved. 

Second Method — Straight-line, No-tension Method. 
— This method we name as above, because the additional as- 
sumptions are indicated by such name. 

We will now make the following additional assumptions — 
(a) All the tensile stress is carried by the reinforcement. 



220 



THE STRENGTH OF MATERIALS 



(6) For the concrete the stress is proportional to the strain 
(c) The area of reinforcement is so small that we may assume 
the stress constant over it. 

Fig. 107 shows the section, strain diagram, and stress 
diagram. 

We will first give the usual treatment which is based upon 
argument from first principles. 

In accordance with our first assumption a vertical plane 
section becomes an inclined plane section a.' b', the neutral 
axis (N.A.) being at the point c. 



Neutral /fxi's 
— ^ #- 



rt 



1 




Fig. 107. — Reinforced Concrete Beams. Method 2. 



What we first require to determine is the position of the 
N.A. 

Now A A.' and d d' represent the maximum strains in the 
concrete and the steel respectively, and since the line a' b' is 
assumed straight, these strains are proportional to their 
distances from the neutral axis. 



. We have 



max. strain in concrete 



n 



max. strain in steel 
but max. strain in concrete 



{d~n] 
c 



(7) 



and max. strain in steel = ^ 



n _ c E, 
• " • (^ -"ti) ~ t ' E, 
. • . nt = mc{d — 7i) 



m c 



(8) 



STRESSES IN BEAMS 221 

, nt 

d-n = 

m c 



_t_ 
m c 



n=- f (9) 



d 

1 + 



mc 

This determines the distance from the N.A. when both c 
and t are known ; but this will not always be the case. If the 
reinforcing bars are of given size, then t will depend on that 
size, and to determine the position of the neutral axis, we 
proceed as follows : The stress diagram shows the distribution 
of stress in the cross section. Since we have assumed the 
stress proportional to the strain, the stress diagram for the 
concrete will be a triangle. It will be seen therefore that the 

mean compressive stress is -^, and since the compression area 

is b n, we see that the total compressive stress is ^ cb n. 

As the stress in the steel is assumed uniform, we get that 
the total tensile stress in the steel is t A^, and if the beam is 
subjected to pure bending these must be equal. 

.-. tA, = ~ cbn (10) 

c 2 A, 
I.e. - = -,- . 
t on 

Comparing this with equation (8) we get 

2 A, n 

b n m{d — n) 
.'.bn^ = 2mA,{d-n) (11) 

.-. bn^ = 2mAyd — 2'mA^n 
.' . b 71^ + 2 m A, . n — 2 m A., d = 0. 

The real solution of the quadratic equation gives 



»-6^{-^W(^ + ^a)} (-) 



222 THE STRENGTH OF ]MATERIALS 

Since all the quantities in this expression are given, this 
fixes the position of the neutral axis. 
We may write this 

n _mA, r / ~2hd_ A 
d bd iV ^ mA, J 





i.e. 


n 

d ^ ''- = 


r m 


u 


1+^ - 
r m 




= Wm^ 


r^ 


+ 2 


mr — in r 


For m 


= 15. 


This gives 
r = 

•007 
•010 
•015 
•020 






n 

d ^ ''^ 

•365 

•417 

•483 
•530 



1 



(13) 



These values are plotted on Fig. 108. 

Moment of Resistance. — We can now find comparatively 
simply the moment of resistance. The resultant compression 
acts at the centre of gravity of its triangle. 

Therefore the distance between the resultant compressions 

and tensions is d — q- 

.•. If C and T represent these resultant compressions and 
tensions, we have that the moment of the couple due to the 
resisting stresses, which is called the moment of resistance, 
is given by 

MR ^c(d - ^^ 

= ^c69^((^-|) (14) 

or, MR = T(d - !? 

= ^A,(cZ-|) (15) 

And this moment of resistance must be equal to the maximum 
bending moment for the loading. 



STRESSES IN BEAMS 



223 



160 




f 




















II 


/ 








1 






















«7 


c/ 


/ 








'50 




















v 


W 1 


!# 




^ 


5^ 


v^ 






















4 


1 


^ 


- 


n 






I40 




















/ 


/) 


1 




r ' 


4 


y 






















// 


/ 




^ 


V^ 






150 


















/ 


4 


/ 




X 


























y 


P 












c\ 


120 
















/ 


// 


/ 








C 


■^ 


\J 
















/ 


J 


ii 


/ 






^ 








110 














/ 


1l 


1 




^ 






















/ 


/ 


/ 


1 


i 




y 












100 










/ 


/ 




■ 


X 










c 


-t 













/ 


/ 




/ / 


\r 










^ 








go 








/ 


/ 




7/ 










^ 
















1 






^, 


// 




y 
















80 






j , 


/ 


/ 




/ 




y 






















1 / 


/ 


/ 


' ■ 


/ 




















70 




/ 


1 


/ 


/ 


^1 


























// 


/ 




^1 


/ 






















bo 




// 


1 


// 




























1 


/ 


/ 


// 


f 
























30 




/ 


/ 


/ 
















, 














7 


1 






















^^ 




"^ 






/ 


k 


1 








.L 


•? 


cJ^ 


X' 












^ 




' 


III 








Cj! 


£> 


^ 
















• • 

__30 




f 


II 




, ^ 






















^ 


^ 






^ 


^ 
























V 


^ZO 




I 


/ 


























(0 
0) 




y 


¥ 




























^ 


^ 10 



































■50 



■40 



^0 



20 



rerceniaae neinjorceme.nr 



?/o 



v£ /4i. /6 



70 



Fig. 108. — Rectangular Reinforced Concrete Beams. 



224 THE STRENGTH OF IMATERIALS 

Numerical Example. — Take the same section as worked by 
the previous jormula {see Fig. 106) ; and take c = 500 lbs. 
per in. 2 

Then equation (12) gives 

_ 1-44 X 15 f /, "72 X 6 X 10 ,) 
""- 6 W^+ 1-44 X 15 ~^i 

= 5*61 inches. 
cZ - 71 = 10 - 5-61 = 4-39 inches. 

Then M.R. or safe B.M. considering the concrete is equal to 

500 X 6 -^^1 f. on , 2^ „.l . „ 
— 2 X 0-61 4-39 + „5'61 m. lbs. 

= "^'^- X 5-61 X 8-13 = 5,700 ft. lbs. nearly. 

Comparing this with the safe B.M. by the first method we see 
that the present is more than three times as much. 

Stress in steel is then equal to y-^--- 

A, (^ - 3 

5700 X 12 _^„ „ . , 

= 1-44 X 8-13 = ^'^^^^'-P"'^^'^ 

Assuming a span of 10 ft., the max. B.M. if the load is 

W X 10 

uniformly distributed is „ ft. lbs. 

W X 10 

••• ^—"^-^ = 5700 .-. W - 4560 lbs. 

This includes the weight of the beam, which is roughly 

10 X 12 X 6 ic,r. lU r.-A 11 

TXi X 150 lbs. = 7oO lbs. 

144 

. • . Safe load uniformly distributed = 4,560 — 750 

= 3,810 lbs. 

It will be seen from the stress in the steel that the area of 
reinforcement is more than it need have been. By combining 
equations (9) and (10) we could have found the value of A, to 
give the stress in the steel, say 16,000 lbs. per sq. in., when the 
compressive stress in the concrete is 500 lbs. per sq. in. 



STRESSES IN BEAMS 



225 



The above formula gives results which are in fairly good 
agreement with tests, and is the one most largely used in 
practice. 

Alternative Treatment for Straight-line, No-Ten- 
sion Method. — The following treatment follows more nearly 
the ordinary method of dealing with beams than the above, 
but it is not nearly so often used in this country. We shall, 
however, find it very useful in the case of beams other than 
rectangular ones with tension reinforcement only and so we 
give it here; its xalue has been scarcely sufficiently ap- 




FiG. 109. 

« 

predated. Fig. 109 shows the section of the beam and 
also the equivalent section. 

We find the position of the neutral axis for the equivalent 
section by the rule given on p. 196 that the total first moment 
of the cross section of the beam about the N.A. is zero. 

71 

.' . b n X j^ = 1st moment of compression area about N.A. 

— mAt{d — %) = 1st moment of equivalent tension area 
about N.A. 

b n^ 
•'• 2^ 



mA,{d — n) =0 (15) 

OT bn^ -i-2 7nA,n -2m A, d = [cf . p. 221] . . (12) 



This is the same relation as before. 
Q 



I 



226 THE STRENGTH OF IVIATERIALS 

Equivalent moment of Inertia of the section 
= Ik = o + m A^ {d — n)^ 

Iv 71 

= o + o {d — n) [from 15 above] 

h 71^ [ J 71 

^ 2 V^-3 
Now M = ^ ^^ (15a) 

71 ^ ' 

[compare ordinary bending formula p 197] 
=^cb7i^ld— „ j 

' ~2n 
-lcnb(d-'^') (10) 

This agrees exactly with our previous result. 
Considering the stress on the reinforcement 

(156) 



M 


= 


m 


{d - 71) 


c 






t 


71 




7n 


{d - 71) 


71 






1 


"'■ d 


~ 




t ' 






1 


7n c 



as before. 



This is used if A^ is not given. 

Case in which Stresses are given and Area of Steel 
has to be found. 

In the case we have n^ ^ 

1+ ' 

w c 
Suppose for instance t ^ 16,000 and c -^ 600 and ?n = 15 

1 ' _ ^^ _ .Oft 

^1 ~ 16,000 ~ 25 

"^ 600 X 15 

.-. n = -36 cZ. 



STRESSES IN BEAMS 227 

Then from equation 10 we can calculate the necessary area 
of reinforcement by the relation 

K _ch n 

K en. 600 X -36 ^.^^^^ 

' ' hd 2t 2 X 16,000 ' 

In this case the moments of resistance given by equations 
(14) and (15) will be equal and 

^. -^ 600 6 X -36 ^(6^ - -12 d) 
M.R. = 2 

.-. SafeB.M. = 95 6^^2 

The coefficient 95 is called the resistance modulus and 
can be plotted in very convenient form as in Fig. 109. These 
curves may be likened to the tables of section moduli for steel 
sections. 

Numerical Example. — A reinforced concrete beam is 
required to carry a bending moment of 240,000 in. lbs. Design 
the section for stresses c = 600, t = 16,000, assuming that the 
breadth of the beam is 10 inches. 

By equation {U) d = ^ ^^-^ 



240,000 
M 95 X 10 
= 15*9 inches. 

.-. A- ^ -00675 
a 

A, = -00675 X 15-9 x 10 

= 1*07 sq. inches. 

Adopt 2 — J"' bars [giving an area 1-2]. 

Third Method — General No-terision Method. — In this 
method we will, as before, assume that all the tensile stress is 
taken by the steel, but we will assume that the stress-strain 
curve for concrete is not straight but some other curve. 



228 



THE STRENGTH OF IVIATERIALS 



In this way we get the stress diagram, Fig. 110, from the 
strain diagram. 

Suppose that its area = k . c . ii and that its centroid is at 
distance y from the top. 

Then, as in equations (8), (9) we get 

{d — n) m .c 



n — 



. n 



d 



1 + 



mc 






/. 


»~ 






A 


n 






d 




, 






d- 


r. 


cL, 


------- 


-•-'- 






r 



Fig. 110. — Reinforced Concrete Beams. Methods. 



Now, since the total compressive stress must be equal to the 
total tensile stress, we have 

tAj = khnc (16) 

{d — n) m A^ 



k n 
. • . kh 11^ ^ m A^{d — n) 
.' . k b 71^ -{- m Aj 71 — m Aj d 

A^m ( 







^ ^ 4:dkb '\ 
'' ^ 2X6 W ^ + m A, ~ ^/ 



or. 



d 



r TTi ( 
2l\ 



r m J 



(H) 
(18) 

(19) 



Then moment of resistance 

= M.R. =tA,{d-y) for tension (20) 

= kb 71 c {d — y) for compression . . (21) 



STRESSES IN BEAMS 229 

Numerical Example with Stress -strain Curve Para- 
bola. — Take the section that we have worked for the 
previous formulae. (See Fig. 106.) 

If the stress-strain curve is a parabola tangential at the 
compression edge we have 

2 







3ri 




» ■frko mTTon cr^/^'firk'n 'w — ^ j / 1 1 . 


6 


. tilt; glVCll OCOulUll fi — jj 1 A / X 1 1 ;? 1 ,AA 

2.-3.6 


,3 


= 5-12" 




.-. d -n = 10 - 5-12 = 4-88 




Safe M.R. for concrete 




= 1 X 6 X 5-12 X 500 (4-88 -f 3-20) in. lbs. 




= 3 X j2 X 5'12 X 500 X 8-08 ft. lbs. 





■} 



=: 6,900 ft. lbs. nearly 
Then stress in reinforcement 

6,900 X 12 H TOAiu 

= ' = r44->^^ir8 = ^'^^^ ^^'- P^^ ^^- ^^- 

It will be seen that this method gives higher values still for 
the safe bending moments. The stress-strain curve for con- 
crete, although nearly parabolic, would not have the vertex of 
the parabola at a stress of 500 lbs. per sq. in. 

From the above we think that it should be clear that there 
is not much difficulty in finding the stress in reinforced concrete 
beams so long as we know accurately the properties of the 
concrete, and are clear as to what assumptions we are making. 

Reinforced Concrete T Beams. — Reinforced concrete 
floors usually consist of reinforced slabs with reinforced 
beams at definite intervals in a longitudinal direction, the 
whole being monolithic. Fig. Ill shows a section of such a 
floor, which may be regarded as a number of T beams. The 



230 



THE STRENGTH OF MATERIALS 



reinforcing bars a in a transverse direction in the slabs 
are arranged as shown to take the tension at the top 
where the bending moment reverses, due to the slabs being 
continuous. 

It is usual to take the effective breadth of the flanges of the 
T beams as less than B — J to | B — because the concrete be- 
tween the beams acts as a short beam in a direction at right 
angles, and so the centre portion is comparatively highly 
stressed for this reason. 

We will now consider the stress in the beam, adopting the 
no-tension, straight-line method. 

Case 1. If d, > n we get the same rules as given in method 
(2) for rectangular beams, 6, being substituted for h. 




Fig. 111. 



Case 2. If 6?^ <[ ?^ we proceed as follows — 
As before we have from a consideration of the strain diagram 

{d — n) m c • 



n = 



n = 



t 



d 

m c 

Now consider the total stress diagram, Fig. 112, i.e. hori- 
zontal lengths of compression figure = compressive stress per 
sq. in. X breadth of beam. 

Now total compressive stress on the section 

= C = area (k d h — h f g) 

c 6, n (6, — b,) XX 
- "2~~ 2 ^ w 



C 



21 



6s n 



{b. 



-b,)x^\ 
n J 



But C = T = ^ A, 



STRESSES IN BEAMS 

{bs -b,.)x^\ 



. ^ A^ = ^\b,n 



n 



231 

(22) 



2 A. 



c _ 

{b,n — -~ — - 

I n J 

2 A^m {d — n) 



n = 



\ 



b,n — 



{bs -b,)x^] 



n J 



n^ 0, n — ^-" '-—' ]■ = 2 A^m{d — n) 

nih, n + 2d, {b, - b,) + ^ {h, . - 5,)| = 2 A,m {d - n) 

V ft J 

i.e.b^n^ + 2n [A, m + d, {b, - b,\ =^2 A,md ^ [b,- b,) d,^ (23) 



A' 



4 



^4— •*" I / 



- * AjC- 



TV 



/9 





Fig. 112. — ^Reinforced T Beams. 

From this quadratic the value of n can be found. 

Then if the centroid of the compressive stress-strain curve 
area is at distance a from the centre of reinforcement 

Safe B.M. = C x « 

Let the centroid of the compressive stress-strain diagram be 
at distance y from the top. 

Now this centroid is the same as the centre of pressure on a 
similar body subjected to fluid pressure, the N.A. being the 
water line. In this case it is easily shown that 

2nd Mt. of area above N.A. about top. 



y 



1st Mt. of area above N.A. about top. 
b,n^ {b, — b,.) x^ 
~3~ "^ 3 
6., n^ {bs — b,) x^ * * ' 

'~Y "^ "~2 



(24) 



This enables us to find a. 



232 



THE STRENGTH OF MATERIALS 



Many \vriters neglect the rib, i.e. neglect the portion f e h 
of the stress diagram, and others further assume y = '4: d, 

(the extreme limits between which y must lie are ' and ~\. 

This avoids the quadratic equation and makes the calculation 
much easier. 

We may put h,. = in equation (23) 



A 




A' 






d 






f 



jd 



TV 



/9 



Fig. 112a. 



We then get at once 



n 



2 m A, ^ + hs ds^ 
2 {m A, + b, dS 



a then becomes equal to d — o ( n / 

^ 3 \ 2 n — ds 

Considering compression we have 
Safe bending moment = C . a 



(25) 



= I j^ d, (2" - '^4 {d - i- (l^ -^f-)] (26) 

2 n I 3\2n — ds /J ^ ' 

Alternative Treatment. — Applying the method of the 



STRESSES IN BEAMS 



233 



equivalent section we have (Fig. 113) a much simpler 
treatment. 

Moment of equivalent section about N.A. = 



n' 



i.e. 6.,^ — (6, — h,) 



{n - d,f 



m A^{d — n) 



This gives the same quadratic as equation (23) 

I ■ = ^|! _ (b^^MilLSzd^ + mAAd-n)K... (27) 

neglecting the rib we should have 



N 



u 






t 



^ 



H 



d.-'TV 



Fig. 113. — ^Reinforced Concrete T Beams. 



This gives as before 



n = 



2 m At ^ -\-h, d^ 
2 {b, ds + m A,) 



Having found I^ we have as before 

cl 

Safe B.M. = — for concrete 
n 



m {d — n) 



for reinforcement 



234 



THE STRENGTH OF MATERIALS 



XuMERiCAL Example of T Beam.— Take the T beam of 
section shown in Fig. 114. In this case we Avill not assume the 
area of reinforcement (A,) to be given, but will calculate it so 
as to give 

c = 600 lbs. -per sq. in. 
t = 6000 lbs. per sq. in. 
m = 15 

d 

m c 
15 



Then we have n = 



1 + 



1 + 



6000 



= 5-4 ins. 



15 + 600 



-^ 


^ 




»^ 






1 






/' 


3 




— /o' — 




' 













Fig. 114. 



. •. From equation (22) 

16,000A, = ^00/5.4^^g_38xl:9:| 
2, I 5*4 J 

.'. A^ = 4'38 sq. ins. 

.•. Adopt, say, 3 bars If'^' diameter. 

Then working by the equivalent moment of Inertia 

I, = 15 X 4-49 X 9-62 + 48 x \^ - ^^ /<i'9' 

= 8,641 

.-. Safe B.M. = ^^^ ^,^'^^^ = 960,000 in. lbs. 

5'4 

For other cases of reinforced concrete beams the reader is 



STRESSES IN BEAMS 



235 



referred to the author's Elementary Principles of Reinforced 
Concrete Construction (Scott Greenwood & Son, London). 



COMBINED BENDING AND DIRECT STRESSES 

If the loadmg on a beam is such as to cause a direct stress 
in addition to bending stresses, then the resultant stresses 
across the section will be obtained by adding together the 
separate stresses. Let b d, Fig. 115, represent the elevation 
of a section of a beam, c being the centroid of the section 




Fig. 115. 

whose area is A and whose compression and tensile moduli 
are Z,. and Z^, d being the compression side and b the tension 
side. 

Then, if the direct force is a thrust Q, there will be a uniform 

compression stress of -^ over the section. If the bending 

moment is equal to M, the maximum compression and tensile 

M M 
stresses due to bending are equal respectively to ^y and ^ . 

Therefore we have 

Resultant maximum compressive stress = /, = ^ + ^ . . (1) 

Resultant maximum tensile stress = /^ = ^ — ^ . . (2) 

The distribution of the combined stresses across the section 



236 



THE STRENGTH OF MATERIALS 



is then as shown in Fig. 115, fh representing the maximum 
compressive stress, and g e the maximum tensile stress. The 
neutral axis then is at the point n, where the stress is zero. 
If the direct force is a pull T instead of a thrust Q, we have 



Resultant maximum tensile stress = f. 
Resultant maximum compressive stress 



T M 



M T 
'' Z,. A ^ ' 



Stresses obtained from Line of Pressure. — If the 

resultant force across the cross section is R, Fig. 116, and the 





Q 
B 


^ 7 


D ^^ 


^^r 


' Lmt 



Fig. 116. 

line of pressure cuts d b produced in l, the load point, then 
resolving R along and perpendicular to the cross section we get 
a shearing force S and a thrust Q. 

In this case M = QxcL = Qxa; 

and if c D ^ d, and c b = d, 

I _ AF 

d, dc 

1 _ AF 

dt df 

where k is the radius of gyration about a line through the 
centroid parallel to the neutral axis. 
.-. We have from equations (1) and (2) 



we have 



Z.=: 



/c 



Q , Q .x d, 

A~^ AF" 



STRESSES IN BEAMS 237 

= 1(^ + ^0 (') 

, ^ Q, .xd, _ Q 
''~ AF A 

Q^fxd, \ 

= AKk- -^ ^^^ 

Or if the resultant normal component is a pull T, equations 
(3) and (4) become 

A = xfi+#) (7) 



AV Ic^ 

A = ^f#-i) (8) 



Position of the Neutral Axis. — The position of the 
neutral axis n can be found as follows — 
Let it be at distance y from c. 

Then stress due to bending = j 

_ Q^^ 
AF 

At this point the stress due to bending is exactly equal to 

the direct stress, 

Q^xy ^ Q 
• • A^2- A 

or a;?/ = P < 

k^ 

The following numerical examples will make the question of 
combined direct and bending stresses clear ; further examples 
will occur in the course of the book. 

Numerical Examples. — (1) A tension rod is a flat bar 8 
inches wide and 1 inch thick : owing to bad fitting, the line of pull, 
instead of passing along the geometrical axis of the bar, lies i of an 
inch to one side of it, in the plane ivhich bisects the thickness of 
the rod. Determine the maximum and minimum stresses set up 
in this bar in a section at right aiigles to the line of pull when the 
pull is 36 tons. 






238 THE STRENGTH OF IMATERIALS 

Show by a sketch the actual distribution of the stress across the 
section. {B.Sc. Loud.) 

In this case the direct stress = -. = „ — , =4-5 tons per sq .in. 

A 8 X 1 1 M 

The B.M. is equal to T x x, and the second moment is equal to 
1_)^ 83 _ 128 
12 " 3 

,, _ I _ 128 1 _ 16 
••^"~A~ 3 ^8~3 

= 4-5(l+^x4xfg) 

3 

= 4*5 X 1 , ,^ = 5*344 tons jjer sq. in. 

T /xd, 

-Mm-' 

13 
= — 4"5 X 1 /- "= ~ 3'6o6 tons per sq. in. 

The distribution of the stress is then as shown in Fig. 117. 

(2) A hollow circular column has a projecting bracket on which 
a load of 1 ton rests. The centre of this load is 2 feet from the 
centre of the column. External diameter of column is 10 inches, 
and thickness 1 inch. What is the maximum compression 
stress? [A.M.I.C.E.) 

In this case A - ^ (10=^ - 8^) = 28*28 

I = J^ (10* - 8*) = 289-8 in. units 
64 ' 

,., 289" 8 1/i o- 
• • ^^" = 28-28 = ^^^ ^^ 

. / _ Q /^i , ^' ^'•' 

1 / 24 X 5 



+ 



28-28 V 10-25 



STRESSES IN BEAMS 



239 



9" -L 



56Tons. 



^ 



,4 

3-6561 



® 



B 



5-344- 



i 56 Tons 




■'^''^^^© 




3-33 



a.5Tons 




Fig. 117. — Combined Bending and Direct Stress. 



240 THE STRENGTH OF MATERIALS 

12-7 

, _Qfxdt 

= '379 ton per sq. in. 



28-28 
The distance of the N.A. from the centre of the section is 



then given by 2/ = — 



= -427 m. 



24 

The distribution of stresses is then as shown in Fig. 117. 

(3) A built-up crane jib is in the form of a curved girder^ and a 
Jiorizontal section near the base is a hollow rectangle. The out- 
side dimensions of this rectangle are 54 and 36 inches, and the 
larger and shorter sides are 1 inch and 2 inches thick respectively. 
Find the maximum tensile and compressive stresses induced in 
the material when a load of 25 tons is suspended from the end of 
the crane, the horizontal distance of the load from the centre of 
the section being 50 feet. Show by a sketch how the intensity 
of stress varies across the section. {B.Sc. Lond.) 

It will be noted that in this question no means are given to 
connect the plates of the rectangle, such means being necessarj^ 
in practice. 

Proceeding as in the previous example, we see that 
A = 2 (72) + 1 (100) - 244 sq. ins. 
. 36 X 543 34 X 503 



^ ~ 12 12 

.^ = il^« = 484-5 

25 / 600 X 27 



118,200 



244 V 484-5 

25 
== ^.. X 34-5 = 3*62 tons per sq. in. 
244 

Fig. 116 shows the manner in which the stresses are dis- 
tributed. 



STRESSES IN BEAMS 241 

* BEAMS WITH OBLIQUE LOADING 

In obtaining our formulae for the stresses in beams, we 
assumed that " the section of the beam is symmetrical about 
an axis through the centroid of the cross section parallel to 
the plane in which bending occurs." 

We saw in dealing with moments of inertia, or second 
moments, that an axis of symmetry is called a principal axis 
of the section. Our assumption, therefore, is equivalent to 
saying that one of the principal axes lies in the plane of loading 
of the beam. 

When such is not the case the loading is said to be oblique 
and we proceed as follows or in the alternative method given 
on p. 244. Draw the momental ellipse for the beam, x x and 
Y Y (Fig. 118) being the principal axes, and let z z be the trace 
of the plane of loading. Then the neutral axis will be the 
diameter of the ellipse conjugate to the plane of loading. The 
plane of bending will be at right angles to the neutral axis. 

This is proved as follows — 

Consider an element of area at the point p of a section 
(Fig. 118), and let p n and p m be drawn perpendicular to the 
plane of loading and neutral axis respectively. Then the 
intensity of stress at p is proportional to p m, the distance 
from the neutral axis, so that if c is a constant we may write 

/p = c X P M. 

. • . The moment of the load over the area about z z is equal 

to /i. X a X P N = C X a X P M X P N. 

Now since z z is the plane of loading, the moment of all the 
stresses over the section about z z must be zero, since the 
couple to the stresses must also be in plane z z. 

.•.2/pXaXPN=0 

i.e. 2 c X a X P M X P N = 

^. e. 2 a . p M X P N = 

But S a . p M . p N is what we have previously called the pro- 
duct moment, and it can be shown that if the product moment 
of an area about two lines is equal to zero, such lines must be 
conjugate diameters of an ellipse. 

R 



242 



THE STRENGTH OF I^LVTERIALS 



Therefore to find the neutral axis draw a chord the diameter 
conjugate to z z. To do this draw a chord parallel to z z and 
bisect it and join c to the point of bisection. 

Xow suppose the radius of gjTation about the N.A. is ^,, ,., 




Fig. 118. — Stresses due to Oblique Loading. 



and dc and dt are the distances from the extreme points of the 

section to the compression and tension sides respectivel}'. 

Then the moduli are 

\h ^ T 

~d^. " ~ X 

dt dt 



Z, 



STRESSES IN BEAMS 243 

Then the maximum compression and tension stresses are 
obtained by the relations 

/=- 
'' Z,, 

^' " z, 

Numerical Example. — A 5'' x 3"" x i'' unequal angle section 
is loaded on the small side with the long leg downward. Find 
the safe bending moment for a stress of 7 tons per square inch. 

From the tables of standard sections we see that for this 
section the maximum and minimum values of the radius of 
gyration are 1*69 and '65 inches, the principal axes being 
at 19i° to the vertical line z z, which is the trace of the plane 
of loading. 

The momental ellipse is now drawn (to twice the scale in 
Fig. 118), the major axis being equal to twice ^,,,, and the 
minor axis equal to twice k,j,j. 

By the construction previously given we get the diameter of 
the ellipse conjugate to z z. This gives the neutral axis. To 
obtain k^.^ draw a tangent to the ellipse parallel to the N.A. 
and draw a line from c perpendicular to this axis. This will 
be found to be "88 inch. Now measure the distances d,- df 
from the neutral axis to the extreme fibres of the section and 
these will be found to be 1*80 and 1*83 inches respectively. 
The area of the section is 3" 75 sq. ins. Therefore we see 

^ 3-75 X -882 

Z,. = tTqa = 1 dI m. units 

^ 3-75 X -882 . 

Ztf = = 1 59 m. units 

i'oo 

. ' . If safe stress = f,- = ft = ^ tons per square inch 

Safe B.M. - 7 x 159 = 1113in.tons (1) 

If we had taken the N.A. at right angles to the plane of 
loading, as in the case of a symmetrical beam, we should have 
had k = 1-60, cZ, = 1*73, and d, = 327. 



244 THE STRENGTH OF MATERIALS 

This would give Z,. = ^'^A^^^p^' = 5-46 in. units 

y 3-75 X 1-602 

A = 0.07 — ^'^^ ^^' units 

.• . Safe B.M. = 7 x 2-94 = 20;58 in. tons (2) 

(In finding the safe B.M. we, of course, consider only the 
least modulus if the working stresses are the same in tension 
and compression.) 

We see from comparing results (I) and (2) that a very large 
error is made by failing to find the true neutral axis. This 
error is very commonly made by practical designers. 

A similar allowance should be made for symmetrical sections 
where one of the principal axes does not coincide with the 
plane of loading. Such cases occur in practice in plate girders 
where the wind is blowing on one side while the load is crossing, 
and in sloping bridges where the cross girders are placed with 
their flanges at the same inclination as the main girders. 

* Alternative Treatment for Oblique Loading*. — In 
some cases it is much simpler to proceed by what is known as 
the " principle of superposition.'' 

Let X be the angle of inclination of the plane of loading to 
the principal axis and let x, y be the co-ordinates referred to 
the principal axes of the point at which the stress is required. 

Then / = ^1^ + McosX.^ ^^^ 

We shall show later that this comes to a simple result in a 
common case. 

Let N N (Fig. 119) be the neutral axis and let n be the 
perpendicular distance of a point P from it ; then /^, = stress 
at p = m .n where m is a constant. 

Then M sin X = component of M about xx = "% f .a .y 

= '^m .ny a {!) 

but w^ps =PR — SR = PR — QT=i/ cos a — a: sin a 
. • . M sin k = m cos a 2 y^a — m sin a% xy . a (2) 

hut^ X y . a = product moment about principal axes = 
. • . M sin A = m cos a 2 2/^ a = m cos a I^^ (3\ 



STRESSES IN BEAMS 



245 



Similarly M cos \ = component of M about y y 
= '^ f a .X 
= S mnx . a 

= %m {y cos a — X sin a) x . a 
= m cos a^xy a — m sin a^x"^ a 




Fig. 119. — Oblique Loading of Beams. 

= o — m sin a I,^. 

= — m sin a I^y 

Dividing (3) by (4) 

tan X = - -^-^t'^-i- 

J. ~ lyv tan A, 

z. e. cot a = ^J 



(4) 



246 THE STRENGTH OF MATERIALS 

This enables us to calculate the position of the neutral axis. 

T^ ,^. 1 ... M sin X — M cos X 
From (3) and (4) m = ^ — = -x • 

i^x cos a 1,^. sm a 

.' . f = m n = m [y cos a — a' sin a) 
= my cos a — mx sin a 
_ M sin X y cos a / — M cos X x sin a 



I^x cos a \ I^j. sin a 

M sin A. . 1/ , M cos X .x ,», 

= — i~- + ~i- — ('^ 

■■-XX •■-VT 

Numerical Exa]viple. — Take, for instance, the obliquely 
loaded column shown in Fig. 120. Loads of 30 and 40 tons 
respectively are transmitted at a and B, the resultant of which is 
a load of 70 tons acting at d. 

The following are the properties of the section — 

A = 28-59 sq. ins. 
k^^ = 4*45 ins. 
k^^ = 3' 39 ins. 
Measurement gives z o x = A = 110"4° 

4 X 5*5 

M sin A. = 70 X o D sin A = 70 x e o = 70 x — _ — 

= 220 in. tons 
(This is the same as 40 x o b) 

M cos A. = 70 X o D cos X = - 70 x d e = 70 x v x 2-72 

= — 81*6 in. tons 
(This is the same as 30 x o a) 

The maximum stress occurs at the top left-hand corner for 
which y = 5'5, X = — 6 ins. 

,,,,., , 220 X 5-5 , 81-6 x 6 

. • . Max. bendmg stress = / = 28-59~^^4^2 + 28o9 x 3392 

= 214 -f 1-49 

= 3'63 tons per sq. in, 

70 
Direct stress = ^^^^^ = 245 tons per sq. in. 

.-. combined stress = 608 tons per sq. in. 



STRESSES IN BEAMS 



247 



Simplified Result in Special Case. — In the case, as the 
above, where the oblique loading is caused by two bending 
moments in the principal axes, M sin X and M cos X will be 
the separate bending moments in the two axes and we thus 
get the following rule — 




/0^x3iK^^'2//f. 



Calculate the stresses at any point for each bending moment 
separately about the corresponding neutral axes ; then the total 
stress for the two bending moments will be the sum of the separate 
stresses. 



CHAPTER IX 

DEFLECTIONS OF BEAMS 

We have found the relation which exists between the 
stresses in a beam and the bending moment; we now want 
to find the relation between the deflections and the bending 
moment. 

Let c c^ Fig. 121, represent a short length of the centroid 
line of a beam, the original curvature of which was negligible, 
and which has become bent to a radius of curvature R. This 
radius R is that which agrees with the very short length c c', 
and is not the same all along the beam. If the assumptions 
that we previously made with regard to the stresses in beams 
still hold, B r and a e are straight lines after bending, and they 
meet at o, the centre of curvature of c c'. Draw b' f' parallel 
to A E. Now consider the segments b b' c' and c c' o. 

Since $ is very small —. — , = - 

^ B c CO 

B b' b' c' d 
or J = = ^^ (1) 

But A b' represents the length of a b before bending occurs 
B b' increase in length 



" ' A b' original length 

But A b' = c c' 



= strain in a b 



B b' / 

,* We have — , = strain in ab = ~, where / is the 
c c E . 

stress along a B. 

248 



DEFLECTIONS OF BEAMS 

.*. Putting this in equation (1) we have 

/ _ ^ 
E ~ R 
E 
R 



249 



or 4 
d 



But we have already shown that 

d 

M 



or 



M = 

/ 



d 



. • . combining these results we have 

J _ M _ E 
cf ~ I ~ R 



(2) 



(3) 




Fig. 121. 

This is the complete relation between the stresses in beams, 
the bending moment, and the radius of curvature. In practice 
we do not so much want to know the radius of curvature at 
various points of a beam, but we require the deflection, and 
so we will next find the relation between radius of curvature 
and deflection, and then find the deflections for various kinds 
of loading. 

Our investigation now divides itself into two parts according 
as we consider it from the graphical or the mathematical 
standpoint, and we will deal with it in this order. 



250 



THE STRENGTH OF IVIATERIALS 



INVESTIGATION FROM GRAPHICAL STANDPOINT* 

Preliminary Note on Curvature. — Let a b (Fig. 122) 
represent any curve, and let p p^ be points on it at a short 
distance s apart. Draw tangents p Q, p^ q^ to meet any base 
line making angles and 0-^ with it. and draw lines perpen- 
dicular to the tangents, then the point of intersection of these 
perpendiculars is the centre of curvature of the short arc p Pj. 




Fig. 122. 



Then the angle subtended by p p^ at the centre will be 
equal to {0 — 6-^). 

. • . if R is the radius of curvature R x (^ — ^j) = s. 

s 



R = 



e - 0^ 



or 



^ - 0^ _ 1 



R 



1 . 



Then - is called the curvature at the given point, or rather 

R 

A A 

the curvature is the value which ~ approaches as 5 

gets smaller and smaller. 

Mohr's Theorem. — Now imagine a b to be a cable 
loaded vertically in any manner, and let the load between 

* The reader may take either the mathematical or the graphical 
reasoning. Each is complete in itself. 



DEFLECTIONS OF BEAMS 251 

/ the points p, p^ be equal to w. Then it follows from 

Sthe laws of graphic statics that the cable takes up the 
shape of the link polygon, for the load system on it, 
(drawn with a polar distance equal to the horizontal pull 
in the cable. 
Now let the tension in the cable at the points P, p^ be T, T^. 
Then the horizontal components of these tensions must be 
equal, since there is no horizontal force on the cable ; let this 
horizontal component be H; the difference between the 
vertical components of the tensions must be equal to w, the 
load between the points. 

. • . We have H = T cos ^ = T^ cos 6^ 

w; = Ti sin ^1 - T sin 

H sin 6^ H sin 

i.e.w= ^ ^ 

cos Oi cos 6 

= H (tan ^1 - tan 0) 

Now if ^1 and are small, as they will be when considering 
beams, we may say tan O^ = ^^ and tan 0=^0 
.'. We have w = H {0^ - 0) 

w _ H (^ 1 - 0) 
' ' s s 

_ H 

B>it - = load per unit length of the cable = say p. 

H 

-R = S (4' 



Now return to the case of the beam 

1^_ M 
R ~ EI 



1 M 
From equation (3) ^ = ^ j (5) 



The quantity E x I depends solely on the shape and 
material of the beam, and is called the " flexural rigidity." 
Then if this flexural rigidity is constant throughout the span, by 



252 THE STRENGTH OF MATERIALS 

comparing statement A and equations (4) and (5) we see that : 
A loaded beam takes up the same shape as an imaginary cable of 
the same span which is loaded with the bending moment curve on 
the beam, and subjected to a horizontal pull equal to the flexural 
rigidity (EI). 

This is Mohr's Theorem, and the deflected form of the 
beam is called the elastic line of the beam. We see, therefore, 
that to obtain the elastic line of a beam our procedure is as 
follows — 

(1) Draw the bending moment curve for the beam. 

(2) Divide this curve up into narrow vertical strips, and set 
down mid-ordinates on a vector line, and take a polar distance 
equal to the flexural rigidity (EI). 

(3) Draw the link polygon for this vector polygon, and 
reduce it to a horizontal base, then this link polygon gives the 
elastic line to a scale which we shall determine later. 

For the present we will assume that the section of the beam 
is uniform along its length, or rather that the flexural rigidity 
is constant. We shall see later how to proceed when such is 
not the case. 

Standard Gases of Deflections. — In certain special 
cases we can calculate the maximum deflections by reasoning 
based on Mohr's Theorem, and we will deal with such cases 
now (Fig. 123). 

(1) Simply Supported Beam with Central Load W.— 
Let AB represent a simply supported beam of span I with a 
central load W. 

Then adb is the B.M. diagram, the maximum ordinate 

being equal to - . Let A^ c^ B^ be the elastic line of the beam ; 

then, according to Mohr's Theorem, the shape of this elastic 
line is the same as that of an imaginary cable of the same span 
loaded with the B.M. curve and subjected to a horizontal pull 
equal to the flexural rigidity. 

Now consider the stability of one half of this cable. It is 
kept in equilibrium by three forces : the horizontal pull H 



DEFLECTIONS OF BEAMS 



253 



at the point c^ ; the resultant load P on half the cable ; and 
the tension T at the point A^. 

Take moments about the point Ai, then we have 

H X 8 - P X ?/ 

' ' H 

In this case P = area of one-half of B.M. diagram 
_l I Wl _ WJ^ 
~ 2 ' 2 ^ "4 ' ~ 16 
y = distance of centroid of shaded triangle from a 




) 


^P 




^^ 




1 ^^ 


T^ 


— 


j[ """"^ 


ITP 


'1 



Fig. 123. — Deflections of simply supported Beams. 



I 

~ 3 
H = EI 




WP "I 
• ^ " 16 ^ 3. EI 


WP 
~ 48 E 1 



(2) Simply Supported Beam with Uniform Load. — Let 
AB represent a simply supported beam of span I, with a 
uniformly distributed load W. 

The B.M. diagram is a parabola, the height being equal to 

Wl 
'^-. Then considering the stability of half the imaginary 

cable, we have as before 



8 =- 



P X «/ 



H 



254 THE STRENGTH OF ]\LVTERIALS 

In this case P = area of one-half of B.M. diagram 

12 W Z _ WJ2 

~ 2 ' 3 8 ~ 24 

51 

H = EI 

. WZ2 51 5WP 



• 24 ' 16 E I 384 E 1 

•(3) Cantilever with ax Isolated Load not at Free 
End. — Let a cantilever of sj)an l (Fig. 124) carr3'ing a load W 
at a pomt at distance I from the fixed end a. 

Then the B.M. diagram is a triangle, a d being equal to 
W /, Ai B^ represents the elastic line of the beam and the 
imaginary cable. In this case we must imagine the load as 
acting upwards. 

The cable is horizontal at a^. 

Take moments round b^, then we have as before 
H X S = P X 2/ 

. . P xy 
. . 6 == jj-^ 



In this case P = 


= area of B.M. curve a c d 




Wl.l 


WP 




2 


2 




I 
2/ = L-3 






H = EI 






• ^ ^^^' (t 


I \ 



In this case it should be noted that the portion of the beam 
beyond the load is straight. 

(4) Cantilever ^\^TH an Isolated Load at Free End. — 
This is the same as the previous case when Z = l. 

WP / L 

. ' . o = 



2EIV 3 

Wl3 
3EI 



DEFLECTIONS OF BEAMS 



255 



(5) Cantilever with Uniform Load from Fixed End 
TO A Point before the Free End. — Let a b be a cantilever 
on span L, and let a load W be uniformly distributed from a to 
a point c, I being the length of a c. 

Then as before 



8 = 



H 



In this case P = area of B.M. curve a c d 



2 



I = 



W?2 

6 




Fig. 124. — Deflections of Cantilevers. 



y 



L — 



I 



H = EI 
WZ2 



I 
^-4 



6EI 

(6) Cantilever with Uniform Load over Whole 
Length. — This is the same as the previous case when ^ = l. 



8=E^(L 



6EI 
Wl2 3l 
* 4 



Wl^ 



6EI 4 8E I 

* (7) Simply supported Beam with Isolated Load 
ANYWHERE. — The reasoning in this case is somewhat long, but 
should not otherwise present any great difficulties. 



256 



THE STRENGTH OF MATERIALS 



The first important point to notice is that the maximum 
deflection will not occur under the load, so that, as it is the 
maximum deflection that we nearly always require, it is of 
very little use to find the deflection directly under the load as 
is commonly done. 

We have seen that the ordinate of the bending moment 
curve or link polygon of a beam is a maximum where the shear 
is zero, so that treating the B.M. curve as a load on the beam, 




the deflection will be a maximum where the shear due to this 

load is zero. 

Let a load W be placed at a point c on a beam A b of span I, 

Fig. 125, c being at distances a, b from A and B. Then a e B 

W ab 
is the B.M. diagram, c e being equal to — j — . The total 

load represented by this B.M. diagram treated as a load will 

x.t- . Wa6 I Wab 

be equal to the area of the A a e b = , x o ^ ~ 2 * 

It acts at the centroid G of the A. 



DEFLECTIONS OF BEAMS 257 

2 
The vertical through this point g is at distance :;^ r c 

o 

from c, F being the mid-point of the beam, so that the 

distance of this centroid from the end b is equal to 

, , 2 / / \ lb (l + b) 

^ + 3 V2 - ^) = 3 + 3 = ^3— 

. • . The reaction at a due to this imaginary load is equal to 

Total load {I + b) _ W ab {I + b) 

1 ^ 3 ~ Tf ' 3 

Now let the deflection be a maximum at the point d at 

distance x from A. 

Then the shear at this point is zero. 

i.e. R, - I . K H = 

Wab fl + b\ X Wbx 



2/ V 3 / 2 I 

a {I +■ b) 



3 ^ 



or ^ ^ V 3 — ^^^ 

The maximum deflection S is then obtained by considering 
the stability of the portion a-^ d of the imaginary cable. 



Then we have as before 8 



P-2/ 



H 

In this case P = area a k h 



y = 

H = 

8 = 



Wbx X Wbx^ 


I 2 
Wb.a{l + b) 

Ql 
2 
3^^ 


21 

Wab{l + b) 
6Z 


2 la {I + b) 

3 V 3 
EI 

Wab{l + b) 2 jail + b) 1 
6/ 3V 3 EI 

W6 (a(l + b)Y 
3EIZ I 3 J 



258 



THE STRENGTH OF MATERIALS 



This can be put into somewhat simpler form for use by 
putting 

al. Then h = {\ - a)L 



a 



Then^ _ W(l -a) ^al {2 - a) l^ 
~ 3EI \ 3 j 

^3~El(^-"M 3" / 



if' (cfnafi^ 




Fig. 126. — Deflections of Beams. 



* (8) Beam uniformly loaded from one End to the 
Centre. 

The B.M. diagram for this loading is given by the curve 
D T G E (Fig. 126), the curve D t g being a parabola tangential 
to the line e g. 

We have first to find where this maximum deflection will 
occur. We do this by the rule that the Maximum Bending 
Moment in a beam occurs at the point where the shear is zero. 
We will treat the diagram d g e therefore as the load on the 
beam. 

If E G be produced to h, the curve d t g will be a parabola 



DEFLECTIONS OF BEAMS 259 

tangential at G, and it is most convenient in the present j)ro- 
blem to consider the B.M. diagram as made up of the difference 
between the triangle D h e and the parabolic segment d g h. 
The first step is to find the imaginary reaction R^, at e. To 
do this we consider the area of the triangle as a force Pg acting 

down its centre of gravity, which is at distance -^ from d. 



Then Pg = area ofADHE = Jdh.de = 



The area of the parabolic segment will be considered as an 
upwardly acting force p^ passing through its centre of gravity 

which will be at distance - from d. 

o 

Then P^ = area dtgh =Jhd.dk 
_ wl^ I _wP 

~ 3^>r8 * 2 ^ 48 

To get the imaginary reaction R^. at e, take moments 
about D. 

Then 
.-. R 



p, 


I 

"3 


pi- 


= 1^. 


.1 


'e ^ 


3 


Pi 

8 ■ 








48 


w 

8 X 


48 


IwP 




384 



(1) 

Suppose that the maximum defiection occurs at a point 
N at distance x from the centre. 

Then the imaginary shearing force S at this point = 
Shear at n = R^ — area n q e + area Q t G 

384 8 2 "^2 '3 

_1 wP wl fP J ^\ wx^ 

~ 384 16 U "^ ' ^ "^ y ^ G~ 

7 wP wP wP X wlx^ wx^ 



384 64 16 16 ' 6 

wP W X^ W P X wl x^ 

^ 384 "^ "6 l6 16" 



or 



260 THE STRENGTH OF MATERIALS 

If this = 0, we have on dividmg through by . and re- 

arranging the terms, 

64: x^ - 24: xH - 24: xl' - P =0 

'^(jf-^'(jf-^'&^' = '> (^> 

This is a cubic equation that cannot be solved by direct 
methods. 
We must proceed by trial as follows — 

If X = 0. left-hand side, which we will call y = — I 
If a- = l y ^ 064 - 24 - 24 - 1 = - 1-576 

If .1- = -05 y = 008 - 06 - 12 - 1 = - -252 

If X =04 y = 0041 - 038 - 96 - 1 = - 006 

If the values of y are plotted against .v it will be found that 
y = for X = '0406 approximately, and for all practical 
purposes we may take x =04. 

Having determined the point of maximum deflection, we 
have next to calculate the value of the deflection S at this 
point . 

We first find the imaginary Bending Moment ^M^ at the 
point X 

M^ = Re (9 -r x) -}- moment of section Q t g 

— Moment of A q ^' e 

""384 -^ ^*^ ' " ~ 2 ^ 3 ^ " 4 

Wl .-AJs ^^^ ^'^^ 

-Q-- X 04 I X ^ X ^ 

= wl^ ;-00964 - -00001 - 00328; 
= -00637 ivl^ 

M, 00637 wl^ '.., 

•••^ = Ei=— EX- ^^^ 

As an interesting comparison, let us suppose that the ^^ hole 

load were spread right over the span. 

5 W P 
Then 8 = ^„-, ^ ^ , according to the \\ell-known formula. 
384 EI ° 



DEFLECTIONS OF BEAMS 26 J 

In this case W = ^ 

5m;^4 _ -00651 wl"^ 
''' ~ 768 El ~" EI '" ^^ 

We see therefore that we shall only make a slight error^- 
which is practically negligible considering the necessary devia- 
tions from theoretical conditions which occur in practice — if 
we treat for deflection purposes the present case as being the 
same as for the same load spread over the whole span, re- 
membering that the maximum deflection occurs at '54: I from 
the unloaded end. 

Graphical Construction for any Loading. — Let a c b 
be the B.M. curve for any given load system. Divide the 
base into a convenient number of equal parts and let e be 
the length of each base segment. The number is such that 
each piece of the B.M. diagram is approximately a rectangle. 
Now set down the mid ordinates of each section diminished 

in the ratio - on a vector line. These ordinates are diminished 
n 

in order to keep the vector diagram of a workable size. 

Now let the space scale be V = x feet, and let the B.M. 

scale be V = y foot tons. Then considering any section of 

the B.M. diagram, say 2, 3, the area of this section is e x mid 

ordinate. Therefore, on given scales, one inch in height of 

mid ordinate, since the area of each segment is proportional 

to the height of the mid ordinate, represents e x x x y square 

ft. tons. Since each portion of the vector line is — of the 

n 

ordinates, the portion 2, 3 of the vector line represents the 

area of its corresponding section of the B.M. diagram to a 

scale V = n X e X X X y square ft. tons. Now calculate 

the length of E I on this scale. This will be too large for 

EI 

practical use, so take a pole p at distance — , where r is some 

convenient whole number. With this pole p, draw the link 
polygon a' c' b\ then this is the elastic line of the beam for 



262 



THE STRENGTH OF MATERIALS 



the given loading, or, more strictly speaking, a' c' b', when 

reduced to a horizontal base, would give the elastic line. 

The scale to which the deflections are to be read is then 

obtained as follows — 

If the polar distance were taken equal to E I, the deflections 

would be to the space scale 1^' = x feet, but as the polar dis- 

. EI X 

tance is , the deflections will be to a scale V =— feet. The 

r r 

following numerical example should clear up the difficulty 

as to scale — 




Fig. 127. — Graphical Construction for Deflections. 

Numerical Example.—^ 16'' x 6'' x 62 lb. rolled steel 
joist of 24 ft. span carries a uniformly distributed load {in- 
cluding its own weight) of 8 tons, and also an isolated load of 
5 tons, at a point 6 ft. from the left-hand support. Find the 
maximum deflection (Fig. 128). 

In this case E = 12,500 tons per sq. inch. 
I =- 725-7 inch units. 
12,500 X 725-7 



EI 



144 



= 62,980 sq. ft. tons. 



First draw the B.M. diagrams for each of the loads, taking 
as linear scale, say V = 4 ft., and for the B.M. scale, say 
V = 20 ft. tons. Now divide the B.M. diagram into a con- 
venient number of equal parts, say 12, and draw the mid 



DEFLECTIONS OF BEAMS 



263 



ordinate of each part, treating these as force lines, then set 
these ordinates down a vector line, 0, 1, 2, etc. . . 12 to a 
reduced scale, say one-fourth for convenience. 



Then 1 in. down the vector line represents 



4 X 4 X 20 



160 sq. ft. tons, because each base element is J in. 

62,980 
160 



E I on this scale 



373-9 ins. 




Majc.Defn.^ '^^ 

Fig. 128. — Example on Deflections. 

393' 7 

This is obviously not convenient, so take — ^x— 

48 



6'$6ins. 



Then 1 in. on the link polygon represents „^ in. deflection. 

The maximum ordinate of the link polygon will be found to 
be '58 in. 

.'. Maximum deflection = '58 x "8 = '46 in. 

Allowance for Deviation of Cross Section. — The 
cases up to the present have all been on the assumption that 
the section is constant, or rather that the Moment of Inertia, 
I, is the same all along the span. If such is not the case, the 
deflection can be found accurately by first altering the B.M. 
curve to make up for the variation in the section as follows — 

Suppose ABC (Fig. 129) is the B.M. curve on any beam 



264 



THE STRENGTH OF MATERIALS 



A D B, and suppose that I„ is the maximum moment of inertia 
or second moment of the section, this occurring at the point 
D. Then take any point along the beam at which B.M. is x Y 

X Y X I 

and moment of inertia I^ and find x y^ so that x y^ == — ^ "• 

Do this for a number of points along the span, and join up 
the points thus obtained, and we get the corrected B.M. curve 
from which the deflections can be found by the construc- 
tion given above. The value I,| is taken in obtaining the 
expression E I for this construction. 

Deflections of Girders of Uniform Strength and 
Constant Depth. — If the cross section of a beam varies so 
that the maximum stresses are constant along the span, 



r --^Correofed B.M- C6(rve. 




then the modulus of the section must vary in the same way 

M . 
as the B.M., and so the ratio y is constant. If the depth of 

M 

the girder is also constant, then the ratio -j- will also be 

constant. 

The corrected B.M. diagram will in this case be a rectangle, 
and the deflection can be found by Mohr's theorem as 
follows — 

As in the several previous cases we have 

F.y 



S = 



EI 



In this case P will be equal to — ^ and V =-7 since the curve 
is a rectangle. 

, Ml2 



DEFLECTIONS OF BEAMS 265 

W L 

In case of uniform loading M = -g— 

• 64 E I 

Wl 

In case of a central load M = — r- 

4 

• ?i - ^^' 

• 3 2 E I 

Another simple proof of this relation will be found on p. 272. 
Further numerical examples will be found at the conclusion 
of this chapter. 

DEFLECTIONS FROM MATHEMATICAL STANDPOINT 

M 1 

From equation (3) -p, ^ = t^ 

Now when R is great, as it will be in this case, we have 
1 _ d^y 

, d^y M 



• dx'-~^l 

'M d X 
ET 

'^idx 



. -~- = slope of beam = / - 

r r 

y = deflection of beam = / / - 



EI 

Now consider the following standard cases (see Figs. 123, 
124). 

(1) Simply Supported Beam with Central Load W. — 
Consider a point Q at distance x from the centre of the beam. 

ThenM = ^ (^-^ 



W/^ \ ^Ix Wa;2 , 



^ ?r— X 



2\2 J 4 4 



/ 



W I x'^ W X 



8 12 



— + Ci a: + c 



266 


THE 


STRENGTH OF 


MATERIALS 


The 


slope is zero when x = 


.-. 


Cy = 0, and the deflection 


is zero 


when X = 


I 

±2 


• 








WP 
' ' 32 


WP ^ 
96 +^^ = ^ 






C2 = 


- W 

48 


P 


Then maximum deflection 


. occurs when x = 






Then S = 


C2 

EI 


WP* 
~ 48 EI 



(2) Simply Supported Beam with Uniform Load. — 

Taking a point as before at distance x from the centre, we 
have 

. M==f(J-.)-|(l-.> 

- ^f^^ - 

~ 2 Vi ■' 



/ 



,^ , wP X wx^ , 
Mdx ^ — g 6~ ^^ 



+ Cg 



as before c^ = 

wP x^ w x^ 
^ "16 2^ 

= when x = 

• _ _u)P wP 
• " • ~ ^2 - g4 3g4 

" ^ I 384 J 384 

Then the maximum deflection occurs when x = 
c. 5wP 5 W P 



8 = 



EI ~ 384 EI 384 EI 



* The minus sign indicates only that the deflection is downward, and 
need not be employed in calculations. 



DEFLECTIONS OF BEAMS 267 

(3) Cantilever with an Isolated Load not at Free 
End. — Take a point q at distance x from load. 
M = - Wx 



.-. Slope X EI = f 



Mdx 
Wx^ 



2 
When x = — I, slope = 

-WP 

— WP 
. • . E I X slope under load = ^ — 

Mdx 

Wx^ WPx 

6 "^ 2 +^V X ^j 

When X = — I, deflection = 

_ WJ3 _ WJ^ _ - WZ^ 
•'• ^2 - g 2 ~ 3 

. • . Deflection under load, where a: = 
c, /-WZ3\ 1 



r r 

deflection = / / 



_ ^2 _ 



X 



.EI V 3 /EI 
Deflection at free end 

= deflection under load + slope under load (l — I) 
_ /-WJ3 WZ2 ^ 1 

~1"^ 2"^''~^^J EI 

W /Z2^ _P\ 
Ell 2 6j 
W Z2 / _ I 



2 E I V 3 

or neglecting the minus sign, which indicates only that the 

deflection is downward, we get 

W /2 / I 

Maximum deflection = 8 = o-^ftt \'^ ~ ^ 

(4) Cantilever with Isolated Load at Free End. — 

This is obtained by putting Z = l in the above case. 



268 THE STRENGTH OF MATERIALS 

(5) Cantilever with Uniform Load from Fixed End 
to a point before Free End. 

In this case M = — 

.-. Slope X EI =^yM(lx 

w x^ , 

= - g- + c, 
When X = — I, slope = 



c, = 



6 
w P 



E I X slope under load = — ^ 

b 

E I X deflection -= / /M d x 



_ wx'^ wP X 
~ ~ 24 + '""6~ + ""2 
When X == I, deflection = 

_wl^ wl^ _ ~ wl"^ 
•*• ^2- 24~~ 6 ~~8^ 
E I X deflection under load, when a: = 

- C2 - g- 

E I X deflection at free end 

— wl^ 
^ — Q~ + slope under load x E I x (l — Z) 

o 

~ -8 +^^~^^ "6 



-_WP fL _ 11 

2 Is 12/ 

wPf _l 



6 V 4 

- w / V / 

= -6"A^-4 
Neglecting — sign Ave have 



WlV _ I 
6 E I V 4 



DEFLECTIONS OF BEAMS 



269 



(6) Cantilever with Uniform Load over Whole 
Length. — This is the same as the previous case when I = l. 

• • ^-6"Eir 4/ 

~ 8 EI 

(7) Simply Supported Beam with Isolated Load 
anywhere. — Let a load W be placed at a point c on a beam 
A B, Fig. 130, of span I, and let it be at distance a I from the 
end A, the distance from the end b being (1 — a) I. 

Then R3 = — ^— = W a ^ 

R.=WJ-l-^I^^ = W(l-a) 



I 



/^^^ 



^. 



® 



a 



i 



C 



(/ -a.)L 



'B 



R, 



Fig. 130. 



Consider a point at distance x from a between a and c. 
Then M^ = R, a; = W (1 - a) x 

.•.Elg = W(l-a)a; (1) 

.^j^^^Wd-a):.^^ 

ax 2 

E 1 2/ = ^— g — ^ — + Ci a; + C2 (3) 

Now consider a point at distance x^ from a between c 
and B. 

Then M^^ = R, x^ - W {x^ - al) 

= W (1 - a)X-^ -W X^ + W al 
^Wal -WaX^ 



E I -7—^, ^Wal -WaX^ 

a Xjf 



(4) 



270 THE STRENGTH OF MATERIALS 

.-.EI ^^^^ Wa Lr, - Wa"^^' + C3 (o) 

EI?/- ^2 g + C3 .Ti + C4 . . . . (6) 

In equation (3) when .t = 0, ^ = 

.-. C2 = 

In equation (6) when x = I, y ^ 

WaP W a ^3 



2 6 



f Co / + c. = 0. 



3 i' -r ^4 



- W a ^3 
••• C4 = ^ C3Z (0 

These two equations representing the elastic line on either 

side of the load have the same slope and the same ordinate 

when 

X = Xi — al 

. • . putting in these values in equations (2) and (0) and equating 

we have 

W (1 - a) a'-P ,^, , Wa3Z2 

— 5^ — ^ — -i- Cl = ^^ ag /g 2~" + ^3 

Ci = -2 + C3 (8) 

Putting in value x = x-^ = a Zin equations (3) and (6) and 
equating we have 

W(l-a)a3Z3 Wa3Z3 W a* Z3 



Q +CiaZ = 2— ^ g— +C3aZ + C 



C^ a Z = g h C3 a Z ^ C3 I 

Wa3/2 Wa/^ , 
Ci a = ^ 3 r C3 (a — i) 

Wa3/, Wa/2 / WaoZ-X, 



3 3 ' V^ 2 

Wa3J2 ^ Wa/2 _ Wa3Z2 W a^ /^ 

3 3 2 "^ 2 

^ V3 ' 6 2y' 

-WaZ2 

g-- (2 - 3a + a^) 

-^^'"^'(1 -a) (2 -a) (9) 



6 



DEFLECTIONS OF BEAMS 271 

. • . equation (3) becomes 

E I2, = Wg-") ^^ _ W aP X (1 _ „) (2 - a) (10) 

Assume a ^ ^j then the maximum deflection occurs between 

A and c. ?/ is a maximum when v^ = 

z. e. when ^^ ^ ^ — (1 — a) (2 — a) = 

, „ l^a{2 - a) 

I. e. when x"^ == ^-^ 

i.e. ^ = I J r ^^ ~ 3 (11) 

Putting this value in equation (10) we have 

°'^^ = 3Vl"|— 3— I (12) 

The deflection under the load is obtained by putting x = al 
in (10). 

rru i?T W (1 - a) a3 Z^ W a^ P ^, 

Then Ely = — ^ ^ -^ — (1 - a) (2 - a) 

^^-^^7--)[a-2-a] 
_ Wa^(L:^) 

• -^ 3 EI ^^^^ 

Deflection of Girder of Uniform Strength with 
Parallel Flanges. — If the section of a girder varies along 

M 

its length so that the stress is constant, then -^ is constant, 

Li 

M 

so that if the depth is also constant -^ is also constant. 

Assuming also that E is also constant we have 

P = p ,- = constant. 



272 



THE STRENGTH OF MATERIALS 



E I 
. • . Wherever ^ is constant, the beam bends to an arc of 

a circle. • 

Let Fig. 131 represent a beam bent to an arc of a circle. 
(N.B. — The beam is shown vertical instead of horizontal for 
convenience of figure.) 

Then, if is the centre of the circle and c d the deflection 
of the beam, we have from the property of the circle 
c D (c o + o d) = a c2 












Fig 


. 131. 




As 


c D is 


very 


small 


we may write 










8 


2R - 


8R 


P 
4 








Now -p = 


M 
EI 












.-. 8 = 


MP 
8 EI 





This result agrees with that obtained on p. 264 by reasoning 
from Mohr's Theorem. 

* Resilience of Bending-. — The work done in bending a 
beam to a given stress may be obtained as follows — 

The work done by a couple in moving through an angle is 
equal to the product of the moment of the couple into the 



DEFLECTIONS OF BEAMS 273 

angle turned through. Therefore, if a short portion of a 
beam subjected to a bending movement M is bent to a slope 

8 i, the work done in bending is ^ ? because M gradually 

increases from to M. 

. ■ . Total work done in bending over whole beam 



= P= • 2 



JNow , 
a X 


1 1 

~R' '•^- R^ 


.'. P : 


^J 2R^^ 


but^: 


M 
"EI 


.-. P - 


= /2EI^^^ 



1 

rate of change of slope 



// the B.M. is constant, and the section is rectangular, then 

2^1 J 2EI 

o 

But M2 = S^ 
d^ 

• 2 . E . ri2 " 

hh^ , h 
Now I = ^2 . ^ = 2 



. p 


~ E' 

6E 


4 / 

2 / 

.V 


hh^ 
12 


X L 



Where V is the volume of the beam. 

P /2 
.'. Resilience = ,, = /„ 
V 6E 

// the load is central and the section is rectangular. — Consider 
one-half of the beam, then x is the distance from the abutment 



274 THE STRENGTH OF IVIATERIALS 

TIT W X 



p rw d X 



f 



2 y 2 E I 

o 

L 

/w^ .T^ d X 
~ J 8EI 

o 

~ 192 E 1 

96 E 1 2 

A- ^r . . WL /I 

JNow M at centre = - . — = , 

4 d 

• • 6. EI. ^2 

_ f ][L 
~ 6 E ' <^^ 

As before I = -r^, d = 



P = 



Resilience = 



12' 2 

/2 46 /i3 L 
6E' 12/^2 

^^ V 

18 E 

P _ /2 

V ~ 18E 



Numerical Examples ox Deflections, etc. 

(1) A girder lias a span of 120 feet, and has to support a 
uniformly distributed load of 1^ tons per foot run. What depth 
must the girder have in the centre if the maximum deflection 
is not to exceed ytdu ^/ ^^^ span ? The maximum stress in the 
flanges is not to exceed 6J tons per sq. in. and E is 12,000 tons 
per sq. in. {B.Sc. Lond.) 

This question is not quite clear, because if the depth is not 
the same throughout, we cannot calculate the deflection 
until we kno\^' the Ava}^ it varies. 



DEFLECTIONS OF BEAMS 275 

We will assume the depth constant — 

^ ,^ W / 11 X 120 X 120 „^ ^ 

JNow at centre M = -^- == -^ 5 it. tons 

o o 

= 27,000 m. tons. 

M 

. • . If maximum stress = 6 J tons per sq. in. since / = ^ 

Li 

ry 27,000. .. 

Z — — -^. — ni. units. 

5 W L^ 
^^^^ ^ = 384EI 

1,200 10 
12 _ 5 X 150 X 120 X 120 x 120 
•'• 10 ~ 384 X 12,0001 ^ '' 

.-. I = 405,000 in. units. 

Now jj = -^ where D = depth. 

2j: _ 2 X 4 05,000 X 6-5 . 
•'• ^ ~ Z " 27,000 '''^• 

= 30x6;5 _ jg.^g ^^ 

This is a greater depth than would be usually adopted in 
practice for a solid web girder. 

(2) A cast-iron water pipe, 10 inches external diameter and 
J inch thick, rests on supports 40 feet apart. Calculate the 
maximum stress in the outer fibre of the material when empty 
and when full of water, also the corresponding deflections. 
{AJI.LC.E.) 

-r TT (D* - d^) IT (104 - 94) 



In this case 



64 64 

= 168-8 in. units. 

Z^\= ^— = 33-76 in. units. 
d 5 



Volume of pipe = ^ (lOO — 81 j x -rj^ = 4-14 cub. feet. 

Volume of water = ^ • tt^ x 40 =- 17*67 cub. feet. 

4 144 



276 THE STRENGTH OF MATERIALS 

.*. Weight of pipe = w ^ ^ ^..^ ^ '832 ton. 
° ^ ^ 2,240 

Weight 01 water = w. = Tr^.r^ = "492 ton. 

® ^ 2,240 

.'. W = w; + 1^1 = 1324 tons (about) 

. • . Max. stress when empty 

M -832 X 40 X 12 , ,„ , 
^Z= 8x33-76 = 1-48 tons per sq. in. 

Max. stress when full = — ^^r = 2' 35 tons per sq. in. 

Taking E as 8000 tons per sq. in. 

5 WL^ 
8 when empty = -334^^! 

_ 5 X -832 X 4 X 40 x 40 x 1 2 x 12 x 12 
~ 384 X 168-8 X 8000 

= -89 inch. 

8 when full = ^^^^r = 141 inches. 

(3) A pole 7nade of mild steel tube, 6 inches diameter and J inch 
thick, is firmly fixed in the ground, the top being 10 feet above 
the ground level. A horizontal pull of 2000 lbs. is applied at a 
point 6 feet from the ground. Find the deflection at the top. 
E = 13,500 tons per square inch. {B.Sc. Lond.) 

In this case I = — ^ — ^. = n. - - 

64 64 

= 32* 9 in. units. 

This is the same as Case (2). 

In this case ^ = 6 ft. L = 10 ft. 

W = 2000 lbs. = f ^i^.^ ton. 
2,240 

. _ 2000 6 X 6 X 12 X 12 x 8 x 12 

• • ° ~ 2,240 ^ 13,500 X 32-9'x'2" 

= 5 inch, nearly. 

(4) What is the least internal radius to which a bar of steel 



DEFLECTIONS OF BEAMS 277 

4 inches wide by f inch thick can he bent so that the maximum 
stress will not exceed 5 tons per square inch ? E = 13,000 tons 
per square inch. (A.M.I.C.E.) 

The general formula for bending is 

/ME 
d ~ I ~ R 

. i _ E 

' ' d R 

^ d^ 
or K = — 7- 

In this case d = distance from N.A. to extreme fibre, 

3 . 

3 13,000 

= 488 inches. 
= 40-7 feet. 

It should be noted that the width of the bar is not necessary 
in this problem . 

The result is the radius of the centre line. 

(5) A cast-iron beam, has a rectangular cross section, the thick- 
ness being 1 inch and the depth of the section 2 inches. It is 
found that a load of 10 cwt. placed in the centre of a 3Q-inch span 
deflects this beam by '11 inch. Through what height would a 
weight of J cwt. have to fall on to the centre of the same span to 
produce a deflection of "30 inch ? {B.Sc. Lond.) 

It takes 10 cwt. to produce a deflection of "11 inch. 

10 X "30 
.'. It would take — yz. to produce a deflection of '30 

inch. 

Now the work done in deflecting a bar when loaded in the 
centre = J W S, 

.*. Work done to produce '30 inch deflection 

1 10 X -30 ^^ . 
= ^ . — 7Y — X 30 m. cwt. 
■" * J. 1 

= -341 ft. cwt. 



78 THE STRENGTH OF MATERIALS 

If h is the height from which the J cwt. falls, work done by it 
= 9 (^ + T9 ) ^^' ^^^-^ because we shall take h as the height 

above the unstrained position of the beam. 
These two amounts of work must be the same, 



have 


K^^'S)-^^ 




h = -682 ^2 ^^0*' 




= 8-18 - -30 inches 




= 7*88 inches. 



CHAPTER X 

COLUMNS, STANCHIONS AND STRUTS 

The question of strength of columns of compression mem- 
bers is of very great importance, and has formed a field of 
discussion and investigation for many years. Interest in the 
subject has recently been aroused by the regrettable failure 
of the Quebec Bridge, and within the next few years many 
investigators will probably direct their energy towards giving 
us further information in this direction. Although the 
subject certainly presents difficulties, much of the confusion 
which is in the minds of many designers is undoubtedly due 
to insufficient grasp of the meaning of the various formulae 
in use. We will endeavour to make this subject quite 
clear by approaching it in the following manner, which was 
suggested by the author in 1908. 

In the design of a tie-bar we use a constant working stress, 
that is to say, the stress does not depend on the shape or the 
length of the tie ; but in struts or compression members 
the working stress depends on the shape and the length and 
the manner in which the ends are fixed. The quantity which 
determines the working stress, and thus the strength of a 
pin-jointed strut, column, or stanchion is equal to 

Length of column _ 

Least radius of gyration about centroid 

This quantity we will call the Buckling Factor of the strut. 

For struts with ends fixed in other ways the buckling factor 

is obtained by dividing the equivalent length of the strut by 

the least radius of gyration. We will show later how the 

equivalent length is obtained. 

279 



280 THE STRENGTH OF MATERIALS 

Slenderness ratio. — Some writers use this term in place of 
buckling factor. The slenderness ratio 

Length 
Least radius of gj^ration about centroid 

but does not take into account the method of fixing the ends, 
and so is not the same as the buckling factor. Two struts of 
the same material and having the same buckling factor may 
carry the same stress, no matter how their ends are fixed : 
this is not the case for two columns with the same slenderness 
ratio. 

The reason w^hy a variable working stress has to be used 
is that struts fail by buckling and not b}^ crushing, unless 
their length is extremely small. If for some reason the 
centre line of a strut is not quite straight or the load comes 
out of centre, there are bending stresses caused in the material, 
and the distortion due to these bending stresses tends to 
increase the eccentricity, and failure may ultimately occur 
due to this reason. 

Strut Formulae. — A large number of formulae, some 
theoretical and some empirical, have been proposed for ob- 
taining the w^orking stress in compression in terms of the 
buckling factor of the strut and of the crushing strength of 
the material. Before these formulae can logically be compared 
we must be careful to see that they are for the same crushing 
strength, and for the same manner of fixing the ends of the 
strut. We will consider the following : — 

(a) Euler's Formula. — This formula is intended for long 
struts in which the direct stress is negligible compared with 
the buckling stress. It is usually given in the following 

form — 

^EI 

^ " L'^ 
where P = the breaking load (not the working load) 
E = Young's modulus. 
I = least moment of inertia. 
L = length of pin- jointed strut. 



COLUMNS, STANCHIONS AND STRUTS 281 

We will now put it into more convenient use for practice 
as follows — 

P , ,. ^ 7r2EAA;2 

.*. -r = breaking stress = — * t^t— 
A ^ AL2 

_ 7r2 E _ 7r2 E 

~ /L\2 ~:f 

Adopting a factor of safety of 5, we get 

xxT 1 • i / breaking stress tt^E 

Working stress = /,, = J' = -— ^ 

^ 5 5c^ 

For mild steel, E = 13,000 tons per sq. in. 

, TT^E 25,600^ 
• '. /j> = -^2, "^ 2 — tons per sq. m. 

For wrought iron, E = 12,500 

24,600 

• ' Ji' ~' 

Similarily for cast iron /^, = 
For timber /^, = 



12,000 
1,600 



Proof of Euler's Formula. — The proof of Euler's 
formula is found by many students to be somewhat difficult 
to follow, as it involves the solution of a differential equation. 
Suppose that a column in some way or other becomes de- 
flected as shown in Fig. 132 (1) . Then there are bending stresses 
induced in it, and the strut will exert a force P on the supports 
tending to straighten itself. Now, if the load on the strut 
is less than P, the strut will straighten, and so is safe ; but 
if the load is greater than P, the strut will continue to deflect, 
and will ultimately break. When the load is equal to P, 
the strut is in unstable equilibrium, and so P is called the 
critical or buckling, or crippling load. 

Consider a point a on the strut. 



282 



THE STRENGTH OF MATERIALS 



The B.M. at A = M, = -Pa:.* 
.Now, if R is the radius of curvature, 

J^ _6Z2^_ M _ - Pa ; 

R ~ dy^~Wl~ EI 

• ' dj^^ ~ET • ^ = say - m2 . a: (1) 

assuming that I is constant, or that the strut is of uniform 
section. 

\ Z 5 4-5 



-JC- 



1 




Fig. 132. — Methods of Fixing Ends of Columns. 

The general solution of this differential equation is 
X — A cos m ?/ + B sin my , 



(2) 



* The negative sign occurs because we take anti-clockwise moments 
to the right as positive, and x in the figure is, on the usual convention, 
negative. If the figure be turned round so that the column is horizontal 
and has a downward deflection it will be seen that according to the rule 
given on p. 121, M.^ is positive and x is negative. Fx therefore is 
negative, and to make the moment positive we must write M^ = — Pa;. 
If we had drawn the column buckled in the other direction, M,^ would 
have been negative and x positive, so that we still would have 
M., = - Pa;. 



COLUMNS, STANCHIONS AND STRUTS 283 
where A and B are constants, which are obtained as follows — 

When y = -^ and -^^— , a: = 

^2 2 

„ . mL , -r, . mL .^. 

. • . = A cos -^ \- B sm -^r- (3) 

^ . — mL , ^ . — mL ,., 

= A cos — ^ h B sm — ^— (4) 

. mL -r, . wL 
= A cos — ^ — 13 sm —^ 

. • . B must = 

. • . a; = A cos my (5) 

When 2/ = 0, a; is infinite, . * . A is not zero 

.p . mL ^ 
. • . II A cos — ^ = 

A 

mL , ^ 
cos — ^ must = 

A 

The general solution for this condition is that , 

mL _ wtt 

• • • w^ = -L2- 

• • EI~ L2 

P = — L^ (6) 

The lowest value of P is given by n = 1, and as this is the 
most important for us, we write the result as 

^--^ (7) 

It should be noted that P is independent of the quantity x, 
so that the force necessary to keep the strut deflected at large 
radius of curvature is the same as that to keep it at a small 
radius, and so if the load is the least amount greater than P 
the strut will go on deflecting, and so break. 

Use of Euler's Formula. — It must be remembered that 
in this formula we have not taken into account the direct 



284 THE STRENGTH OF MATERIALS 

compression stress on the strut. If the safe stress given by 
Euler's formula is greater than the safe compressive stress 
for very short lengths of the material, then obviously we should 
not use Euler's result. Thus, if Euler for mild steel gives //, 
greater than 6 tons per sq. in. we should use 6 tons per sq. in. 
Another way of using it is as follows — 

Safe load = -- ,^- 
5 i^ 

_p_2EI 

.-. I required = ^ -^ 
A required = P 

We have thus the least area and moment of inertia that 
the section must have and can so choose a suitable section 
from tables. 

Method of Fixing Ends — Equivalent Length of 
Strut. — In the above working we have considered the ends 
as pin-jointed. If the ends are fixed in any other way we 
must take as the length of the strut the length of the equi- 
valent pin-jointed strut; this we will call the equivalent 
length of the strut. 

Now consider the following methods of fixing the ends (see 
Fig. 132). 

(1) Pin Joints at Each End. — This is the standard case. 

(2) Both Ends Fixed in Position and Direction. — In 
this case the buckled form is as shown in the figure, and b c 
is the equivalent length, i.e. a pin- jointed strut of length 
B c is as strong as the fixed strut. 

. * . in this case equivalent length of strut = „- 

Buckling factor = c = ^ , 

(3) Both Ends Fixed in Direction only. — The buckled 
form in this case is as shown in the figure. On comparing 



COLUMNS, STANCHIONS AND STRUTS 



285 



with Case 1, it will be seen that the portion b c is equivalent 
to one-half the strut in Case 1, and so in this case, 

L 

since B c = A 

Li 

equivalent length of strut = L 

L 
I 



. Buckling factor == c 



(4) One End Fixed in Direction and Position, other 
End Pin- jointed. — It will be clear from the figure that in 

this case 

2 L 
equivalent length of strut = — ^— 

2L 



• . Buckling factor = c = 



Zk 



(5) One End Fixed in Direction and Position, other 
End Free. — In this case 

equivalent length of strut = 2 L 

2L 
k 



. ' . Buckling factor 



c = 



Summary of Values of Buckling Factors 





Case 1. 


Case 2. 


Case 3. 


Case 4. 


Case 5. 


Buckling factor 
= c 


L 

k 


L 

2k 


L 

k 


2L 

dk 


2L 

k 



These values should be used in Euler's and the other 
formulae involving the buckling factor. 

{b) Rankine's Formula. — This formula is sometimes 
called the Gordon-Rankine formula, and is of the form 

1 + a . c^ 



/.= 



286 THE STRENGTH OF MATERIALS 

Where 

/, = safe compressive stress for very short lengths of the 

material 
a = a constant depending on the material 
c == buckling factor of the strut 
ii, = working stress per sq. in. for the strut. 

The following values of a may be taken according to 
different authorities — 

Mild steel a = kftf^ to 7.7^K7^, /, = 6 tons per sq. in. 
Wrought iron a = ^^^ to -g^^^, /„ =4 „ „ „ 

Cast iron « = 2500 *%"io' ^'^ ^ " " " 

Timber a = ^qoo' ^' ^ *^ " " " 

In each case we prefer to use the higher value of the constant a. 

There is a very large amount of variation in the values of 
the constants as given by various authorities, and in com- 
paring the above with those given by others, the reader 
should be careful to compare the safe stresses given with the 
above figures with the safe stresses given b}^ others, because 
the value of /^ also varies in the various forms of the formula 
and thus, although the constants may be different, the re- 
sulting safe stress may be nearly the same. Care must also 
be taken to see whether pin-jointed or fixed ends are taken as 
the standard case. 

Construction of Rankine's Formula.— Rankine's formula 
may be looked upon as a corrected form of Euler's. 

If c is very small, i. e. if the strut is very short, the term 
a c^ is negligible, and so we get f^, = /,. 

This is, of course, the result which we ought to obtain. 

If c is great, ^. e. if the strut is very long, the term a c^ will 
be so great that 1 may be neglected in comparison with it, 
and so we get 



COLUMNS, STANCHIONS AND STRUTS 287 
This will give the same result as Euler if 

a 5 

. .„ 1 TT^E 25,600 , ^„„ 
^. e. if - = -^-r = v, — = 4,267 
a Djc o 

Although some writers state that constants obtained in 
this manner agree with experimental results, the constants 
are not usually calculated theoretically in this way, but are 
obtained from experiments. 

It is believed that the figures recommended above will 
agree well with the best practice. 

It is interesting to note that in one form of Rankine's formula, 
giving the breaking or crippling stress, viz. 

p / 

^ + Hi) 

/ is the stress at the elastic limit. 

In an earlier chapter we pointed out the desirability of 
obtaining the working stresses from elastic limit, i.e. basing 
the factor of safety on the elastic limit. 

An interesting and important paper by Mr. C. P. Buchanan, 
in Engineering News, December 26, 1907 — published after the 
Quebec Bridge disaster — gives the results of tests on full- 
size built-up columns such as are actually used in bridge 
practice. The tests extend over a period of fourteen years, 
and show that even for the short columns the buckling or 
crippling stress is not more than 90 per cent, of the tensile 
yield point (see p. 4). 

We thus see that in columns as actually used in practice, 
the buckling stress is certainly not more than the elastic 
limit stress, and so the only reasonable factor of safety is that 
based on the elastic limit. 

(c) Straight Line Formula. — These empirical formulae 
are used principally in America, and give very good approxi- 
mations for rough working. They are of the form 

= /, (1 - e . c) 



288 THE STRENGTH OF IVIATERIALS 

Where f^, and /, are as before 

e = a constant depending on the material. 

The following values of e may be taken — 
For mild steel e = -0053 

,, wrought iron e = "OOoS 
,, cast iron e = "008 

„ timber e = -0083 

As in Rankine's formula the values of constants vary con- 
siderably according to different authorities. 

{d) Johnson's Parabolic Formula. — This is also an 
empirical'formula devised to agree with Euler for long lengths, 
and to agree with the ordinar}^ compression strength for short 
lengths. It is of the form 

= /.(!-?• c^) 
(/ is a constant of such value as to make the curve of /,, 
plotted against c tangential to Euler, and the curve is used 
up to the point where it meets the Euler curve. 
The following values may be taken for g — 

For mild steel g = -000057 
„ wrought iron g = -000039 
,, cast iron g = 00016 

(e) Gordon's Formula. — This formula is often confused 
with Rankine's, and was used largely for some time, but it 
is now quickly going out of use in favour of the Rankine 
formula. This is probably due to the fact that designers are 
now more used to making calculations involving the radius of 
gyration, a quantity which practical men have usually looked 
upon with suspicion. Now that tables are published giving 
k for most sections, it is as easy to use as the diameter d. 

Gordon's formula is of the form 

, L _. 



COLUMNS, STANCHIONS AND STRUTS 289 

Where /,., /^„ and L have their usual meaning. 

j is constant depending on the material and on the shape of 

the section, 
d is the least diameter or breadth of the section. 

The objection to this formula as compared with Rankine's 
lies in the fact that one has to use different constants for 
different shapes of section for the same material. Otherwise 
it is very similar to Rankine's. 

The following values for j may be taken, f, being the same 
as in Rankine — 



Shape of Section. 


j 


Mild Steel. , ^frwf,^* 1 ^^^* ^°^' Timber. 


Solid circle 

Hollow circle .... 
L, T, H, etc. .... 
Built-up sections . . . 
Rectangle (solid) . . . 


1 

370 

1 

600 

1 
300 

1 

400 

1 

500 


1 

500 

1 

800 

1 

400 

1 

550 

1 

700 


1 
110 

1 

180 

1 

90 

1 

120 


1 
125 

1 
200 

1 
100 

1 

160 



(/) Fidler's Formula. — The reader is referred to Fidler's 
Bridge Construction for a very complete analysis of the strut 
problem. 

The formula which Mr. Fidler obtains gives the breaking 
stress, and is 

Minimum breaking stress = / + B + V(7^KF^2 m f B 



m 



Where / = ultimate pure compressive strength of material 



E 



R = Euler's breaking stress 



m — a constant of average value 1'2. 



u 



290 



THE STRENGTH OF MATERIALS 



The following values of f,,, the safe stress in tons per sq. in. 
for struts, are suggested by Fidler and are used by some 
authorities — 



L 

k 


Mild Steel. 


Wrought Iron. 


Cast Iron. i 


Pin ends. 


Fixed ends. 


Pin ends. ' Pixed ends. 

1 


Pin ends. 


Fixed ends. 1 


20 


5-20 


5-29 


3-92 


3-99 


8^07 


8-65 


40 


4-76 


5-09 


3-64 


3-89 


5-68 


7-56 ! 


60 


4-02 


4-83 


3-17 


3-73 


3-35 


6-10 


80 


3-15 


4-45 


2-60 


3-48 


1-96 


4-68 


100 


2-40 


4-00 


2-03 


317 


1^29 


3-35 


120 


1-83 


3-46 


1-57 


2-82 


•93 


2-37 


140 


1-42 


2-96 


1-24 


2^48 


•70 


1-78 


160 


1-13 


2-51 


•98 


2-14 


•56 


r40 


180 


•91 


213 


•80 


1-84 


•43 


114 



Lilly's Formula. — Professor Lilly of Dublin has devised 
formulae for columns which allow for secondary flexure or 
wrinkling in the column,* and may be regarded as a modifica- 
tion or correction of Rankine's formula. 

It is of the form 

1 + w . - + ac^ 

V 

Where tn is a constant depending on the shape of section 
t is the thickness of the metal in the section 

a is -^-^. (Compare p. 287.) 



7)1 



5N/,. 
E '■ 



the values of N being given for some 



sections in Fig. 132a. 

Use of Strut Formulae.— Fig. 133 shows curves of f,, for 
mild steel for various values of the buckling factor according 
to the first four formulae. It is advisable to draw such a 
curve to a good scale, choosing one of the formulae — say 

* See Engineering, January 10, 1908, or a more complete paper and 
bibliography in Proc. Am. Soc. C.E. Vol. LXXVI (1913); also The 
Design of Plate Girders and Columns (Chapman & Hall, Ltd.). 



COLUMNS, STANCHIONS AND STRUTS 291 



Rankine — with a = np,crr\\ such curve can then be used 
whenever the value of /^, is required. 



N. 50 




iV. 60 






iV^. 120 




N.IQ 



Fig. 132a.— Lilly's Column Formula. 

It will be remembered that /^, gives the safe stress per sq. 
in. for struts with central loads. If the loads are eccentric we 
must proceed as described later. 

Then if A = area of section of strut, 

Safe load = P, = /^^ . A. 

If, as often occurs in practice, we are given the load but 



292 



THE STRENGTH OF MATERIALS 



have not designed the section, so that we do not know the 
buckling factor, we can often get a rough idea by taking a 
trial value of ^ equal to about f /,, i. e. 4 tons per sq. in. for 
steel, and finding the area requisite for this stress. This will 
give us an idea of the area required, and we can choose a 
section with roughly this area, and see by finding its buckling 
factor what is the safe load on it. 




ISO 



Buckling Factor = C- 

Fig. 133. — Curves for various Column or Strut Formulae. 
(Mild Steel) 

Many of the leading constructional steelwork firms publish 
tables of safe loads on various struts. Having previously 
checked one or two to see that these firms work with similar 
formulae, we can choose a suitable section for our case, and 
then apply our formula and see if such section is satisfactory. 



REINFORCED CONCRETE COLUMNS 
Short Columns Centrally Loaded. — We have shown on 
p. 38 that the safe load in a column in which buckling is 



COLUMNS, STANCHIONS AND STRUTS 



293 



negligible (the length being less than 15 times the least 
diameter, and the notation being modified) is given by 

P = c (A, + m A,) 

Gross-binding of Reinforcement. — In addition to the 
longitudinal reinforcementj some force of binding is necessary 
to keep the bars at the requisite distance apart. This is due 
to the following reason — 

Suppose that a reinforced column with bars a b, c d,. 
Fig. 133a, be compressed; then, quite apart from any buckling 



u 



R \ C 




B I D 



■^"T 



Fig. 133a. 



of the whole column, the column will bulge out somewhat 
as shown, and the reinforcing bars will buckle because the 

value of ^ or the buckling factor for them will be large. If 

we bind the reinforcing bars together, as shown diagrammati- 
cally, so that they cannot buckle, the column will not bulge 
to anything like the same extent, and so will be considerably 
strengthened. From a large number of experiments M. 
Considere found that the best results are obtained when 
spiral coils are placed round the reinforcing bars at distances 
apart equal to i to ~o of the diameter of the coil. 



294 THE STRENGTH OF MATERIALS 

M. Considere suggested the following allowance for the coils 
in the strength of the column — 

Let A/, be the equivalent area of longitudinal reinforce- 

volume of metal in coils \ 
length of column 

Then safe load = c (A, + m A, + 2 4 m A') 

Long Columns Centrally Loaded. — Some authorities 

use Euler's formula apj)lied to the homogeneous section, 

viz. — 

IT' 



. ,, . , ., / . A volume of metal m coils \ 
ment of the spiral coils [i.e. A,, = , ,, ,, , 

^ \ length 01 column / 



Safe stress = fj, 



^E 

2 



5 c 



c being the buckling factor. In obtaining c the radius of 
gyration of the equivalent homogeneous section (see p. 183) 
is used, 

"i: 



where A = A^ -f- m A,. 

I^. = equivalent second moment 

= L + (m — 1) A, r^ for section shown in Fig. 133a. 

Y being the moment of the section apart from the reinforce- 
ment, 

I^, = ^- (- [m — 1) A, r^ for circle 

b h^ 
= --— -f (m — 1) A, r^ for rectangle 

Then safe load = fj, x (A, + m A,) 
Rankine's formula can also be used in the form 

_ _ 5 00 

^ 8000 

Braced Columns, Struts, and Stanchions. — Struts 
are often formed of roUed sections such as beams and channels 
braced together by diagonal bracing or plates. The strut 
that failed in the Quebec Bridge was a braced strut, and the 
report of the Commission states that there is not yet sufficient 



COLUMNS, STANCHIONS AND STRUTS 295 



information for the design of such struts for very heavy 
loads.* For ordinary comparatively light work, however, 




. ^-N 


X 


y-^. 








Y 


— D 


---^ 


^ 


5- 
Y 


^^^^ 




-* 


P 


^ 






I 


v^-^ 






^^ 


i> 



o 



o 



a — 



o 



o 



o 



o 



o 



o 



s 




Fig. 134. — Columns with Open Webs. 

braced struts such as shown in Fig. 134 are commonly used 
but the diagonal bracing should preferably have one rivet 

* See Illinois University Bulletin, No. 44, G. Talbot and Moore, for 
an experimental investigation of the subject. 



290 THE STRENGTH OF MATERIALS 

passing through the two diagonal bows as in the top of Fig. 
135 instead of two rivets as shown. The unbraced length 
of one of the beams or channels must be such that the load 
per sq. in. on them is not more than the safe stress for them 
considered as struts. We can get an idea of the maximum 
unbraced length as follows — 

Let c = buckling factor of whole strut 

,, ki = least radius of gyration of one channel or beam 

,, P = total load carried b}^ strut 

,, 2 A = total area of strut 

,, S = maximum unbraced length of channel or beam. 

Then, using Euler's Formula, ^^-v- = '^ = ^ 



Each channel or beam carries ^ load 

stress 






5S2 ~ S2 
B _ B yfci^ 

• • C2 ~ S^ 

. • . S = k-^c 

r. ' Equivalent length of strut L 

yjT Since c = = 

' Least radius of gryation of whole strut k 

S _h, 
L k 

L k 

.' . ^ = least number of panels = r- 

_ Least radius of gryation of whole strut 
Least radius of gyration of channel or beam 

As a rule a spacing of 2 to 3 times the breadth B or 30° to 
45° inclination of the diagonals will be found to be satisfactory, 
and in practice would be adopted, unless the calculation 
required them to be less. 

The strength of the strut in this case is calculated as if the 
section consisted of the two channels or beams held at the 
requisite distance apart. See worked Example No. 4. 

Relative Values of Different Bracings. — Professor H. F. 



COLUMNS, STANCHIONS AND STRUTS 297 

Moore, of Illinois University,* has given the results shown in 
Fig. 135 of the " flexual efficiency " of various forms of bracing. 
This flexual efficiency is the ratio of the calculated fibre stress 
to that obtained by measuring the strain, the calculation 
being made on the assumption that the braced section behaves 
as an integral one. The tests were made by ordinary cross 
bending and not as columns, but the comparative results 
may be taken as representing the relative values of the different 



ICO 












































^^ 


^ 




— 
















w /^ '/Tk yv' /?-■ /O 














N 


\ 














■:mm>f^' 












\ 


^ 








60 


'^t^^m 


""" , 


'^ 


^^^ 














^_ 








1 










^ 


^ 














4U 




-— 


















^^ 


a 




^^ 


_^ 















^ 


^- 


--^ 


. 














■mam] 






"^ 





















i 


> 


4 


\- 


« 


? 


I 


i 


IC 


) 


r 


> 


(4 



Stress in thousands of pounds per sq. in. 

Fig. 135. — Efficiency of Bracing. 

kinds of bracing for column purposes. Particular attention 
is directed to the great advantage that single diagonal bracing 
with single rivets (the third from the top) has over that with 
the separate rivets (bottom) in which heavy secondary stresses 
occur. 

Least Radius of Gyration. — The least radius of gyration 
will be about or at right angles to an axis of symmetry if 
there be one, so that in this case we need only calculate k for 

* See a paper h»y Professor A. H. Basquin, Proc. Western Society of 
Engineers, 1914, on " The Design of Columns." This is one of the best 
papers which have been published on the subject. 



298 THE STRENGTH OF MATERIALS 

the axis of symmetry and at right angles to it. If there is 
no such axis we should proceed as indicated on p. 172. 

Examples on Struts, etc., with Central Loads. — The 
following numerical examples should make the question of 
the design of struts, etc., clear. 

(1) A 10'' X C X 42 Standard I beam of mild steel is used 
as a stanchion, the length being 16 ft. and one end being fixed 
and one end pin- jointed. Find the safe load for it to carry. 

From the table of standard sections we see 

A = 12-35 sq. ins. 

Least k = 1"36 

T, , ,. „ , equivalent length 2L 

.'. UucKilins; lactor = c = - ^ «^ = or 

° l\5o 6 k 

2 X 16 X 12 o^ o 1. + 
== Q ^ 1 .QA = 94-2 about 

6 X 1 OO 

.•. Safe stress = f,, = 0^02" ^^sing Rankine's formula 

1 + — 
^ 6000 

= 2"43 tons per sq. in. 



1 + 1-47 

.-. Safe load = 12-35 x 2-43 = 30 tons. 

(2) A solid cast-iron column, 6 inches in diameter and 15 feet 
long, is fixed at the lower end and carries a load at its free upper 
end. Calculate the load the column will safely carry, assuming 
a reasonable factor of safety. {B.Sc. Lond.) 

In this case k = . =1-5'' 
4 

Equivalent length = 2 L = 30 

_ equivalent length _ 30 X^12 
•'• ^ ^ k~ ~ V5 

= 240 

7 
.-. Safe stress per sq. m. = /,, = 240">r240 

^ ^ 1,800 

_ 7__ 
" 1 + 32 
= -212 ton per sq. in. 



COLUMNS, STANCHIONS AND STRUTS 299 

.-. Safe load = '212 x - 4" ^ = 6 tons. 

4 



According to Enler /,, = ^ ^ — o 



7r2 E _ 12,000 
5 c^ c^ 

12,000_ _ 
240 X 240 



.-. Safe load = -'- = 5-88 tons. 

4 

(3) A steel rolled joist is used as a strut with built-in ends, 
the length of the strut being 15 feet. Find, from the data given 
below, the cross section of the joist, if it has to support a com- 
pressive load of 40 tons with a factor of safety o/ 4. 

(a) The total depth of the cross section of the joist is twice the 
width of the flanges, and the thickness of metal is to be 
f of the width of the flanges, 
{b) The crushing strength of a short strut of this quality of 
steel is 24 tons per square inch. 

(c) The constant in Bankine formula is ^^ ^^^. {B.Sc. 
Loud.) 

In this problem we must first find the breaking stress from 
the formula. In this case we do not use the equivalent length 
of the strut because the constant is given for fixed ends. 

^ . . . 24 



x^x^^cirvj-iig i 


3UJ.COO - 


^ 36,000 V A;/ 




• Sflfp ' 


stress ^ 
A 


breaking stress 


6 


Now let 


= area of section 


1 /Ly 

36,000 VA:/ 


and let 


B 


= breadth of flange 




then 


2B 


= depth of beam 






B 

8 


= thickness of metal. 




Then 


A 


^2B^xB^B(,^_ 


2B\ 
8 ) 






B2 7 B2 15 B2 
~ 4 "^ 32 ~ 32 


•4687 B2 



300 THE STRENGTH OF MATERIALS 

The least radius of gyration will be about an axis perpen- 
dicular to the flanges. 

-r B B3 7B /B\3 

Then I=4-12+4^r2-U; ' 

B* 7 B* 

— 4- — 02111 B* 

- 48 + 12 X 2,048 " ^^^^^ ^ 

3_I _ 02111 B^^ 
' • " " A " -4687 B^ ^^^ ^ 

.-. Safe stress = ^?= ' 



A ~ 15 X 12 X 15 X 12 

"^ 36,000 X 045 B2 
40 6 



•4687 B2 1 , 9 
1 + 



45 B2 



1 + ^4TbO = 40--^^^^^^ 
= -15 X -4687 B2 

.-. B*(-15 X -4687 X -45) - -45 B2 - 9 - 

316B4 - 45B2 - 900 = 

The solution of this quadratic gives 

B2 = 18-2 nearly 
say B = 4J 

. • . Adopt a joist 10'' x 5'' with metal ^" thick. 
We could work this problem roughly by the given rule, as 
follows — 



Take /, 


= ^ X 6 = 4 


.-. A 


= . = 10 sq. ms 


15 B2 
• 32 


= 10 


B2 


10 X 32 64 
~ 15 "3 



B = -^- = 4-62, say 5'' 

V3 

(4) A steel column in a bridge-truss has pin-jointed ends and 
is 26 feet long. It consists of two standard W x S^' X 28-21 lb. 



COLUMNS, STANCHIONS AND STRUTS 301 

channels 'placed 4 J inches apart. Find a safe load for the section. 

(See Fig. 134.) 

On looking up the tables, we see that for a 10'' x 3 J" x 28-21 

lb. channel, 

A = 8-296 

"'max. = O I I 
^min. ^^ *Jy4: 

Dist. of C. G. from edge = P = -933 
Then for whole strut 
k,,. = 3-77 

= 3-1832 + -9942 
. • . k,, = 3-33 

Length _ 26 x 12 



c = 



Least radius of gyration 3*33 

93-6 



• / ^ 


6 


• • '^ 93-6 X 93-6 
"^ 6000 


2-46 


= 2-44 




. Safe load = 2-44 x area 




= 2-44 X 2 X 8-296 




= 40-4 say 40 tons. 





STRUTS WITH ECCENTRIC LOADING 

Simple Approximate Method. — If the thrust in the 
strut is out of the centre, i.e. where there is bending 
moment as well as direct thrust on the strut, we cannot use 
the same rules for design as in the ordinary case. 

In such case we may obtain approximate results by pro- 
ceeding as follows — 

Let the load P be at distance e from the centroid of the 
cross section, then M = P . e (Fig. 136). 

Case 1. Very Short Struts. — If the length is less than 



302 



THE STRENGTH OF MATERIALS 



10 times the least diameter of the strut, the stresses are 
obtained as shown on p. 237. 



i. e. /, 


A ^Z. 


f, 


M P 
Z, A 


In this case /,. 


V Vx 

" K^ Z,. 




P P . e . d, 

~ A "^ Ak'' 




A V^ ^ k^ ) 


P 
•'• A 


ed. 




ZPlufr.o IZy^a 



Y 



Ccniroia L 



D 



■Cff^ — *■ 



-<+ 



B 



F 




Fig. 136. — Columns with Eccentric Loads, 

This gives the safe load P for a compressive stress /,. This 
ease is fully dealt with in Chap. VIII. 

Case 2, Struts Longer than 10 J)iameters. — In this 
case we must make some allowance for buckling tendencies, 
and ^\'e may proceed as follows — 

As in the previous case we have 

Combined compressive stress = . (l + , 2 j 

Now in this case this compressive stress should not be more 



COLUMNS, STANCHIONS AND STRUTS 303 

than the safe stress x>er sq. in. obtained by considering the 
buckling formulae, 

A , , ed,. 

' Safe central load on strut 
i.e. Safe eccentric load on strut = 7— — — ^ 

(1 + i 

where e = eccentricity of load 

dr = distance from centroid to edge of section nearest 

load 
k = radius of gyration about axis perpendicular to the 
plane containing the centroid and the load. 
This formula may be put into a form which is sometimes 
more useful as follows — 

Let P^ be the central load, which is equivalent to the 
eccentric load P. 

Then Px = P (1 + p' 

In this formula e should be taken as the effective eccentricity, 

/M 
i.e. ( p where M is the Bending Moment on the column), the 

value of e shown on the drawings is only true when the 
column is free at the top. For other cases see articles by 
the author in the Architect's and Builder's Journal, March 3 
and 31, 1915. 

P 

Then p may be called the eccentricity factor for the strut. 

In using this formula it should be noted that it is Avorked 
on the assumption that the buckling will take place in the 
plane of the figure, and so the value of k for the strut in this 
direction should be used in finding the safe central load. 

If the safe eccentric load according to this formula comes 
more than the safe central load for the least value of k (this 



304 THE STRENGTH OF MATERIALS 

can of course only occur when the least value of k is about 
the axis d b), the lower value should be used. 

Stanchions -with Web and Flange Connections. — 
The loads on stanchions are often communicated from girders 
connected by cleats, etc., to the web or flange of the stanchion. 
If such connections come on one side only, or if the loads 
communicated from the two sides are not equal, the load will 
not be central, and allowance for the eccentricity should be 
made. 

Numerical Example. — A mild steel stanchion 30 feet long 
and with ends fixed has the section shown in Fig. 136. Find 
the safe central load and also the safe loads communicated at 
the points b and c. 

In this case A = 40*59 sq. ins. 

^xx ^^ 4'o7 ,, ,, 

fCyy = o'41 ,, ,, 

.*. Buckling factor = c = ^^ = „ ^—.^ = 52-8 

* 2k 2 X 3*41 

*** ~ 1 I ^^:3 X 52-8 " 1-464 ~ 
6000 
.-. Safe central load = 40-59 x 4'10 = 166 tons nearly. 

Load at c— e = 225 + 575 = 2-725 

.-. 4 = 6'' 

ed,. 2-725 x 6 



P 3-412 



1-41 



1 (KCK 

. • . Safe eccentric load at c = ,~ , , -. , == 69 tons nearly. 

1 + 1 41 "^ 

Load at B. — We must now first calculate ^ as if k^^ were 

30 X 12 
minimum radius of gyration, i. e. c = „ v „_ = 36-9. 

&j ' 2 X 4-87 

• f , ^ ^ — = 4-89 

• • '" 36- X 36 -9 

"^ 6000 

X = 6" 

d,. = 6" 

X (Z, _ 6 X 6 , _^. 

•'* T2 ~ 4-872 



COLUMNS, STANCHIONS AND STRUTS 305 

c, n ^ • T J ^ 4-89 X 40-59 

. • . bate eccentric load at b = 7 ^ ko ~ 

_ 4-89^ X 40-59 
~ 2-52 

= 77-7 tons nearly. 

In this case the eccentricity factors for c and b are 2-41 
and = 2-14 respectively. 

A rough rule sometimes adopted is to use 2 J and IJ as 
eccentricity factors for flange and web connections respec- 
tively, but such rule is not reliable. It is more nearly true 
for I beams used as stanchions than for built-up sections. 

Alternative Approximate Method. — Where the bending 
stress is large compared with the direct stress it seems reason- 
able to allow that instead of the previous treatment we shall 
subtract the bending stress from the value of /, used in the 
strut formula. 

The compressive bending stress for an effective eccentricity 

' . using the Rankine formula we shall have 

, P e d, 

i =^ = ~ ~^^ 
'"A 1 + ac2 

P / , , c, ^ ed,. 



AV ' ¥^ 

Cast-iron Struts Eccentrically Loaded. — In dealing 
with cast-iron struts with eccentric loads it must be remem- 
bered that they will probably fail by tension. 

The safe load P from the tension standpoint 

X dt , 



306 



THE STRENGTH OF :\L\TERIALS 



when /, is the safe tensile stress, and this should be compared 
with the safe load from the compression standpoint, and the 
lower value adopted. 

* Modified Euler Theory for Eccentric Loading. — 
In this case we have, Fig. 137, 




or putting 



Fig 


. 137. 


— Eccent 


ric Loading of Columns 




P (e 


+ x) = 


M = - 


Elf" 
dy^ 




•*• 


d'-x 
dy'~ 

m = 


_ P ( 

EI^ 

/P 

\ E I 


X + e) 






d^-x 
dy'~ 


— m {x 


+ e) 




' '^j 



The general solution of this is 

(a- 4- e) = A cos my -\- 3 cos m y 
* Cf. p. 282, equation (1). 



(1) = 



COLUMNS, STANCHIONS AND STRUTS 307 

Since x = for y ^ ± ^, B = as before 

. • . (x + e) = A cos my (2) 

. • ^ when .X' = 0, e = A cos ^^ 

. m L 

.'. A = e sec ^ 

m L 

. • . X = e sec — ^— . cos my — e 

= e (sec — ^ . cos my — 1 j (3) 

At point o where y = eccentricity = x„ + e = e^ 

/ m L ,\ , m L 

= e ( sec — ^ I j + e = e sec -„ - 

='^"^2^7© ••■•'^^ 

where f^ "^ ~\ 

. • . stress at o = . I 1 + -\-o~ 

A\ F 

= (putting ^ = c) A. (l + '^f sec ^ ^g (6) 

Now let = r^ ^ '— and call it the Eulerian angle. 

2 Ve ^ 

Then, stress at o =/,.(! + ,2 ^^^ ^ ) C^) 

Values of sec 6 are given in the table on p. 308, taken from 

Professor Basquin's paper * previously referred to. 

c. c. J ■ 1 ^ Safe central load 
. • . bate eccentric load = — — ^ 

1 + ^-^ sec 

This can only be used by trial if the load is not given. 
* Proc. Western Society of Engineers, 1914. 



308 



THE STRENGTH OF MATERIALS 











Buckling Factor c. 








50 


60 


70 


80 


90 


100 


110 


120 130 


140 


5 


105 ] 


L-08 


Ml 


1-15 


1-20 


1-25 


1-32 


1-40 1-60 


1-62 


6 


1-07 


LIO 


114 


M8 


1-24 


1-31 


1-40 


1-51 1-63 


1-82 


7 


1-08 


111 


1-16 


1-22 


1-29 


1-38 


1-50 


1-64 1-83 


2-08 


8 


109 ] 


L-14 


M9 


1-26 


1-35 


1-46 


1-60 


1-79 205 


2-41 


9 


1-10 : 


L-15 


1-22 


1-30 


1-41 


1-54 


1-72 


1-97 2-32 


2-85 


10 


1-12 


1-17 


1-25 


1-34 


1-47 


1-63 


1-86 


2-18 2-65 


3-46 ' 


11 


M3 


L-19 


1-28 


1-39 


1-54 


1-74 


2-02 


2-44 3-12 


4-38 


12 


M4 ] 


L'21 


1-31 


1-43 


1-61 


1-85 


2-22 


2-56 3-74 


5-88 i 


13 


M5 


1-23 


1-34 


1-49 


1-69 


1-98 


2-42 


3-16 4-63 


8-69 


14 


1-17 : 


L-25 


1-37 


1-54 


1-77 


2-12 


2-68 


3-69 602 


16-4 


^15 


1-18 : 


1-28 


1-41 


1-60 


1-87 


2-29 


2-99 


4-40 8-46 


172, 1 


§16 


M9 ■ 


1-30 


1-45 


1-66 


1-97 


2-47 


3-38 


5-43 14-4 




o 17 


1-21 


1-32 


1-49 


1-72 


2-09 


2-69 


3-87 


704 39-0 




U 18 


1-22 


1-35 


1-53 


1-79 


2-21 


2-95 


4-51 


9-86 




g.19 


1-24 


L-37 


1-57 


1-87 


2-36 


3-25 


5-39 


16-4 




1^0 


1-25 


1-40 


1-62 


1-95 


2-51 


3-62 


6-65 


47-4 




0) 

f^21 


1-27 


1-43 


1-66 


2-04 


2-69 


407 


8-63 






-§22 


1-28 ■ 


L-45 


1-71 


2-13 


2-90 


4-65 


12-3 






§23 


1-30 


1-48 


1-77 


2-24 


3-13 


5-40 


20-9 






(§24 


1-31 


1-51 


1-82 


2-35 


3-40 


6-41 


65-9 






Si 26 


1-33 


1-54 


1-88 


2-47 


3-73 


7-87 








1-35 


1-57 


1-94 


2-61 


4-10 


10-1 








§27 


1-37 


1-61 


2-00 


2-76 


4-57 


13-8 








^28 
^29 


1-38 


1-64 


2-08 


2-93 


5-15 


22-9 








1-40 


1-68 


2-15 


311 


5-84 


57-3 








30 


1-42 


1-72 


2-23 


3-32 


6-79 










31 


1-44 


1-75 


2-32 


3-56 


8-04 










05 32 


1-46 


1-79 


2-41 


3-83 


9-86 










2 33 


1-48 


1-84 


2-51 


4-14 


12-7 










3q34 


1-50 


L-88 


2-61 


4-50 


17-3 










■■^35 


1-52 


L-92 


2-73 


4-93 


28-7 










^ 36 


1-54 


1-97 


2-83 


5-43 


68-7 


Table of Eulerian Secants 1 


§)37 
g38 


1-56 
1-59 


2-02 
2-07 


2-98 
313 


6-02 
6-80 






sec 


(^V4) 




^39 
^40 


1-61 


2-13 


3-29 


7-73 






E = 


30,000,000. 




1-63 


2-18 


3-46 


9-07 












41 


1 1-66 


2-24 


3-66 


10-8 








7 




42 ' 1-68 


2-31 


■3-87 


13-3 






jVIultipliers of , „' 


1 


43 1-71 


2-37 


4-11 


17-2 








fC- 




44 1-74 


2-44 


4-38 


24-6 






in Expression for 




45 1-76 


2-51 


4-68 


430 






Maximum Stress. 




46 1-79 


2-59 


503 


172 












47 1-82 


2-67 


5-42 














48 1-85 


2-76 


5-88 














49 1-88 


2-85 


6-42 














50 1-91 


2-95 


7-06 















COLUMNS, STANCHIONS AND STRUTS 309 

Numerical Examples. — (1) Take the same case as dealt with 
on p. 304 for the load atB. 

We found load = 77' 7 tons = - — .^ ' lbs. per sq. in. 

= 4,300 lbs. per sq. in. nearly 
. • . sec = 1*03 about 
The effect of this upon the result is negligible. 
(2) Find the stress produced in the column of question (4), 
p. 300, if the load is 1'' out of centre, in the weak direction, 

^ P 40x2,240 ip^cAnii. 
Here -r = — ^ ^ ' — = 10,800 lbs. per sq. m. 
A 8-296 ^ ^ 

c = 93*6 .*. sec = 1*6 approx. (from table) 

1 X 1-6 X 5-75^ 



.-. stress = 10,800 (1 + 3.332 

= 10,800 (1-83) 

= 19,800 lbs. per sq. in. nearly 
= 8'8 tons per sq. in. 
The approximate method would have given 

stress = 10,800 (l + ^^^3!^) = 10,800 x 1-52 

= 16,400 lbs. per sq. in. nearly 
= 7'3 tons per sq. in. 

Johnson's Formula for Eccentric Loading. — This 
formula, due to Professor Johnson, is obtained by adding the 
additional eccentricity due to the deflection and is 

maximum stress in column = x "^ WT~2 (^) 



P P 1<^^__ 

A . ,,_PL2FA 
^^^ 10 EPA 

Pfl+-- '^' 

I ^ V^ 10 E A k^ 



■"• lOE 



A I kHl- ^ ^' 






(9) 



310 THE STRENGTH OF MATERIALS 

A somewhat more correct but similar formula can be ob- 
tained by regarding the bending moment as uniform; this 

gives a deflection (see p. 264) = ^ = |^' = WE^A^B 



f, e c 

"SE" 



. effective eccentricity = 8 + e-=e(l+ '-^") 

V 8 E/ 

. I 1- ^ P(8 + e) d. 

. . bendmg stress = ^ .!. ' 



AF V" ' 8E 
- j^ (approx.) 

af(i-/^-; 

/, d. e 



^ ^^ 8 E 



Total stress = direct stress + bending stress 
=: f t f,^d,e 

^ ^^ 8E 



b{i /''^^J (1^) 



8E, 
Professor Morley * obtained the same result by the expansion 

, . 0'\ 5e\ 610'' 

sec^ = l + 2,+^ + -gj-+ ... 

Taking first the two terms as an approximation 

sec ^ . / ' ' ^ I + ^ L 
2 \E 8 E 

.*. Equation (7) becomes 

Total stress = /., |l + ^^' (l + g'-^)j 

* Theory of Structures (Longmans). 



CHAPTER XI 

TORSION AND TWISTING OF SHAFTS 

We have seen that in a beam the bending moment is resisted 
by a complex series of stresses in tension and compression 
which vary in intensity at different points in the depth of 
the beam. In the case of shafts we have twisting moment in 
place of the bending moment and the stresses are pure shear 
stresses which vary in intensity at different distances from the 
centre of the shaft. 

Stresses in a Shaft Coupling. — -As an introduction to 





Fig. 138. — Stresses in Coupling Bolts. 



the subject consider the case of the stresses in the bolts of 
the flange coupling shown in Fig. 138. 

Suppose that a twisting moment or torque T is being trans- 
mitted through the coupling from the shaft A to shaft B. 
The shaft b has a resistance to its motion which produces a 
reverse torque numerically equal to T and the effect of these 
opposite torques upon the coupling is a tendency to shear the 

bolts. Suppose that the bolts are at the same distance from 

311 



312 THE STRENGTH OF MATERIALS 

the centre and are so small that the shear stress over them may 
be regarded as constant and that they are equal in area and 
equally stressed. 

Then S = shearing force on each bolt 

.' . Taking moments about the axis of the shaft we have 
Resisting Torque = -^. x r 

.•.we have T = (1) 

or if n is the number of bolts 

T = "i-^"^'-- '■ (2) 

If the shaft is transmitting a horse-power, H.P., and is 
rotating at N revolutions per minute, the work done per 
revolution is 2 tt T so that the work done per minute is 2 tt N T 

, ^ H.P. X 33,000 „^ „ 
we have 1 = ^r ^-r ^^- l^^s. 

H.P. X 33,000 X 12 . ,, 
= 2^N ""■ "'^ '^) 

In calculations upon the strength of shafting it is always 
desirable to work in in. lbs., and taking the unit of 1000 lbs. 
as a " kip " we can work in in. kips to save writing a number 
of O's and thus running the risk of error in dealing with large 
numbers. 

In the case of the shaft coupling that we have considered it 
should be pointed out that we have made a great assumption 
in regarding the bolts as being equally stressed, because if, 
say, three of them are loose fits and the fourth is a good fit, the 
fourth one will carry all the load ; the same point holds with 
ordinary riveted joints. In practice, however, a number of 
bolts are always used and each is always regarded as carrying 



TORSION AND TWISTING OF SHAFTS 313 

its proportion of the load, and the best way to meet the 
difficulty seems to be to have the workmanship as good as 
possible. 

General Case of Torque on Groups of Bolts or 
Rivets. — Suppose that we have any number of bolts or rivets 
of different areas and at different radii from the axis o about 
which a twisting action may be considered as taking place ; in 
Fig. 139 we have shown three such bolts. Then if we imagine 
a slight rotational movement of one part of the joint or 
coupling about the point o relatively to the other it will be 




Fig. 139. 



seen that the movements of the centres of the bolts will be 
proportional to their radii r^, r^, r^, etc., and therefore the strain 
and consequently the stress on any bolt is proportional to its 
distance from the a:»is. 

Let <s„ be the shear stress at unit distance from the axis and 
let Aj, Ag, Ag, etc., be the areas of the various bolts. 

We then have, if Si, s^, and s^, etc., are the shear stresses on 
the various bolts and S^, Sg, S3, etc., the forces 

^1 = s, ri 
^2 ^^ "^'t ^2 

<^3 "= <5(( T^ 



314 THE STRENGTH OF MATERIALS 



^ 



we have 




'otal Torque = 


= T = Si r^ - S2 ^2 ^ S3 r. - . . . 




= s^ Ai rj ^ 5, A2 ro - <S3 A3 r, -^ . 




^ <S„ ^1 /*! -7- 5„ ^1-2 /"o" "^ S^ ^3 7*3" — . 




= «„ ;Ai r/^ + A, r,3 + A3 r,2 -^ . . 




= -^^ 2 A, /• 2 



(4) 

But 2 Ai /-i" is the polar moment of inertia of the group 

of bolts, etc. = I,. 

T 
••• "5" = T (5) 

Numerical Examples o>' Couplings axd Joints. — (1) 
Find the diameter of holts necessary in a coupling which transmits 
120 H.P. at 75 revolutions per minute. The diameter of the 
circle of the holt centres is lOJ inches (i. e. r = o"25 inches) and 
the coupling has 6 holts. The stress allowed is 2 J tons per sq. in. 

In this case from equation (3) 

120 X 33,000 X 12 

1 = — c^ 1-r^ iri • lbs. 

2 IT X 75 

= 100,000 in. lbs. nearly. 

Also from equation (2) 

T = '. X o'zi^ m. tons 

4 

6x2-5x7rX6Z2x2,240 ^ ^^ . „ 

= — — . x5-25m.lbs. 

4 

= 139,000 d~ in. lbs. nearly. 

, _ 100,000 

•'• ~ 139,000 

, /i6o,oo() ^^ . , 

^ = Vr3970'00 = '^^^^^-^^"^^^^'' 

. • . Ado pt bolts Y ^^ diameter. 

(2) ExA3iPLE OF Cleat. — We ivill now take the case shown in 
Fig. 140 of the cleat given in the Handbook of Messrs. Dorman, 
Long d; Co., Ltd., for a 16 in. hy 6 in. standard I beam with a 
minimum span of 18 ft., the rivets being of f in. diameter. 

The safe uniformly distributed load given for this span and 



TORSION AND TWISTING OF SHAFTS 315 

beam is 25 tons, so that the reaction at each end will be 

25 

^ = 12-5 tons, and half of this will be carried by each angle. 

or the load P will be 6-25 tons. 




<^--4-> 



- 2i'-^Z^ 






McVi 



625 






Fig. 140. 



First find the position of the centre of gravity of the rivets. 
It is clearly on the horizontal line through the rivet 3, and its 
distance from the line 1, 3, 5 is obtained by moments thus — 

5 ^ = 2 X 21 

^ 4-5 ^ . 

1. e. a = -zr = 9 m. 
5 



316 THE STRENGTH OF MATERIALS 

Then we tabulate the dimensions as follows — 



No. of Rivet. 


r 


r2 


1 
2 
3 
4 
5 


4-58 

2-62 

-90 

2-62 

4-58 


21-06 

6-88 

-81 

6-88 

21-06 


2 7-2 = 56-69 



Pa; _ 6^25 X 3^5 
•*• * ~ 2>2 - 56^69 

= -348 ton. 

The moment load will be a maximum on rivets 1 and 5 
because they are farthest from x, and will be equal to 
Tg = -348 X 4-58 = 1-59 tons. 



The direct load W on these rivets 



6-25 



= 1-25 tons. 



Therefore resultant load = R5 = 2*20 tons. [See Fig. 140.] 
Now bearing area of a |-in. rivet in a |-in. plate 

9 



_3 3 

~4 ^ 8 



32 



sq. m. 



2-20 X 32 
Bearing stress on rivet = ^ = 7'82 tons per gq. in. 

Area of a |-in. rivet in section ^ -r x \~^] = '442 



.4 

2-20 
. • . Shear stress on rivet = ^^rj^ = 4-98 tons per sq. in. 

The above calculation shows that the rivets are stressed just 
about up to what is commonly taken as a safe working stress for 
rivets in shear, viz. 5 tons per sq. in. The importance of 
allowing for the eccentricity of the stress wiU be clear from 
this example, because the resultant maximum stress on the 
rivets comes to nearly twice the value which would have been 
found if the eccentricity had not been taken into account. 

Torsion of a Circular Shaft. — Suppose that a circular 
shaft of length I and diameter d is subjected to a twisting 



TORSION AND TWISTING OF SHAFTS 317 

moment or torque T (Fig. 141). To preserve the equilibrium 
of the shaft, equal and opposite torques must act at the two 
ends and each normal section of the shaft will be subjected to 
strains which will allow a slight twisting motion of such section 
without causing it to bend or warp out of its plane. 

We will therefore make the following assumptions in 
developing our theory of torsion — 

(1) That plane normal sections of the shaft remain plane after 

twisting. 

(2) That stress is proportional to strain, i. e. all the stresses 

are within the elastic limit. 
A line a b on the circumference of the shaft initially parallel 
to the axis becomes bent to the form A b' as a result of the 




Fig. 141. — Torsion. 



ajoplication of the torque, the end a being regarded for 

convenience as fixed. Then B b' may be called the arc of 

torsion and it subtends at the centre o an angle called the 

angle of torsion. This angle of torsion may be regarded as a 

torsional deflection. 

If we imagine the shaft divided up into a number of very 

small equal slices it follows that since each slice is exactly 

like every other slice the angle of twist in each sHce wiU be 

equal, and if we regard each slice for convenience as of unit 

length we have 





Angle of twist per unit length of shaft = 



I 



Now consider the slice contained between sections x x and 
Y Y, Fig. 142, the thickness x being regarded as very small, 
and consider a square abed of side x at distance r from the 



318 



THE STRENGTH OF :\L\TERIALS 



centre. In our figure 6 c is appreciably curved, but that is 
onlv because we cannot draw the fio^ure clearly without making 
X of appreciable size. 

The result of the twisting action is to make abed take up 
the form a h' c' d, this being the typical form (cf. Fig. 1) indi- 
cating pure shear strain, and the initial shear strain /? is given 
by the relation 



X 



Y 



-^ 



JU^ 



X -^ 




Y 

Fig. U2.— Torsion of Shafts. 



Shear stress — shear modulus x unital shear strain 
= y3G 



X 



G 



(1) 



Xow the angle hob' — angle of twist in length x 

= angle of twist per unit length x x 
_6x_ 
I 

and bb' = ar c = radius x angle = — ^ 

rOx .G 

xl 
rOG 
I 



. • . In (1) shear stress 



(2) 



TORSION AND TWISTING OF SHAFTS 319 

This gives the important result that : The shear stress at 
any point in a shaft is proportional to the distance of that point 
from the centre. 

The shear stress is therefore the same at all points on circles 
concentric with the shaft and the variation of shear stress 
is indicated by the triangle o eg, the stress at the extreme 
fibre being s and that at any other radius r being equal to 

, _s r 
^ ^ R 

Now consider a very small element of area a at a point 

p in the section at distance r from the centre, the area being so 

small that the stress over it is constant. 

r OG 
Then from (2) stress on element = — v- 

• . Force on element = S = stress x area 

rOG 

= -r- • " 

. * . Moment about o of force on element 

= S X r 

rOG 
= -^~.a.r 

G „ 

= -r ■"■' 

. ' . Total moment about the axis of all the forces on the 
section 

= Sum of separate moments 

= > — , - . ar^ 

6 G sr\ 

= —-.— ^ ar^ because 0, G, and I are constant 

P 
= —j~ X Polar moment of Inertia of section 

^GI,. 



I 
But the total moment about the axis of all the forces on 
the section must be equal to the twisting moment, so that 
we get 

T = -^4^" (3) 



320 THE STRENGTH OF :NL\TERL.\LS 

From (2) we get ^^^^i = t^ 

and from (3) we get — = ^ 

^ I, I 

By combining these results we obtain the following com- 
plete relation for torsion which should be compared with the 
corresponding relation on p. 249 for the bending of beams. 

stress T ^G 

^^ = ip=i~ <■*> 

In practical calculations we are usually concerned with the 

maximum shear stress -s 

5 T 



.-. we write t? — y 



T R ... 

J (o) 



By analogy with the method of dealing with the section 
modulus of a beam we call -^ the jjoJar modulus Z . 

T 
We thus get -s = ^ 

otT = f^Z (6) 

Gases in which the Formulae are Applicable. — These 
formulae are based upon the assumption that plane sections 
remain plane after twisting and this is true only for circular 
sections (soUd and hollow) ; for shafts of other section the 
approximate formulae given on p. 333 may be used. 

Solid Ciectxab Sectio>'. — This is of course by far the 
most common case of shafting which occiu^, and in this case 



. Z = 

= -196.5 03 (7 



32 

77 D^ _ D _ - D3 
32 2 ~ 16 

■' 16 



TORSION AND TWISTING OF SHAFTS 321 
Hollow Circular Sectiox. — In this case, Fig. 143 {a), 



Z, = 



i. e.T = 



16 Di 

16 Di 



(8) 



When the metal is Yerj thin, Fig. 143 (6), we have 





Fig. 143. 



I^ = I D* _ (D _ ^j4 j and if t is so smaU that squares 

and higher powers of jt may be neglected we may -^Tite 

(D - ^)4 = D* - 4D3^ 
_ - BH 

I^, - D2 f 



*• e- Z,, = J) 



2 

5 7rD2i 



(9) 



Alternative Derivation of Formula for Solid Shaft. — 

The formulae (7) can also be derived as follows, and although 
we recommend the previous method as the more satisfactory 



322 



THE STRENGTH OF MATERIALS 



"we find by experience that some students find the alternative 
method more easy to follow. It is similar to the method wliich 
we have ahead}' used for beams in some cases (see pp. 220-222). 
Consider a very small sector a o b, Fig. 144, of the circle, so 
small that we may consider a o b as being practically a very 
narrow triangle. Set up a D, B c, to represent the maximum 
shear stress s and complete the pjTamid a b c d o as sho\^Tti. 
Then if this pyramid be considered as di\'ided up into 
a number of slices as indicated, the volume of each shce 
= area of piece of sector x stress on it = load on each 




Fig. 144. 



piece of sector; the volume of pyramid therefore re- 
jDresents the whole load acting on the sector, and to find the 
moment about the point o of the force on the sector we 
regard the volume of the pjTamid as acting at the centre of 

gravity of the p^Tamid which is at distance --. - from o. 

Now the volume of a pyramid = J area of base x height 



ab 



= g Ra 



A D X O A 
1 



R 



5aR2 



TORSION AND TWISTING OF SHAFTS 323 

3R 



Moment about o = Volume x 



4 



3 4 

5 a R^ 



2 
Now in the whole section there will be — of these sectors, 

a • 

because the whole circumference subtends an angle 2 tt at the 
centre. 

.*. Total moment about o of all the forces on the 

5aR3 2 7r 

section = — -. — X — 

4 a 

~ 2 ~ 16 

But this total moment must be equal to the torque T 

. rp _ SttW 
' 16 

This agrees with our previous result (equation (7)). 

Horse-Power Transmitted by Shafting. 

We have seen on p. 312 that 

^ H.P. X 33,000 X 12 . „ 
T = ^^ m. lbs. 

H.P. X 33,000 X 12 . ^ 
= — c. — TVT — cti^at. m. tons. 

2 TT N X 2,240 

. • , putting this into equation (7) we have 
sjjrW _ H.P. X 33,000 X 12 
16 ~ 2 TT N 

H.P. X 33,000 X 12 X 16 



D3 = 



27r2N<S 



^/H P 

This gives D == 523 ^ ^ — for s in tons per in.^ 



VHP 

= 68'4 a/ q^^ — * for s in lbs. per in.^ 

In using this formula it should be remembered that N is to 
be in revolutions per minute, the resulting diameter being in 
inches. 



324 THE STRENGTH OF MATERIALS 



Taking s = 7,500 lbs. per sq. in. this gives 

N 



D = 3-5 J 



This is a very convenient formula for use. In practice s is 
often taken rather less than 7,500 lbs. per sq. in., the table 
given on p. 335 being often used; this is based upon s == 6,800 
and is arranged to give round numbers for the 10 in. shaft. 

Calculation of Angle of Twist. 

From equation (3) 

G Ij, 

This, of course, will be in radians. 
For solid shafts this gives 

= -^ — ^. radians (10) 

= -Q7p4 degrees (11) 

Comparison between Solid and Hollow Shafts. — Sup- 
pose that we have two shafts — one solid and of diameter D, 
and the other hollow and of external diameter D and internal 

diameter^. 

Then I^, for solid shaft = ^^ 

Then I,, for hollow shaft = ^ JD* - (^^^^ 

_ TT 15D4 
"" 32 ' 16 

For the solid shaft we have 

^1 " 16 

and for the hollow shaft we have 

^2 - 16 ^ 32 • 2 

15 7^^ 

" ^ ^ 16 ■ 16 
^2 _ 15 
•■• Ti 16 



TORSION AND TWISTING OF SHAFTS 325 

The hollow shaft will therefore transmit }f of the torque 
of the solid shaft and therefore |;? of the horse -power. 

The area of the hollow shaft will be v ( D^ — , ) = — ~ — > 

4 \ 4 / 4 

7rD2 

and of the solid shaft it will be — - — ; so that the area, and 

4 

therefore the weight per given length of the hollow shaft, is 

f of the corresponding value for the solid shaft. 

Summing up our results, therefore, we may say that " the 

hollow shaft has a weight of J that of the solid shaft and 

transmits yf of the horse-power; so that weight for weight 

15 4 

the hollow shaft will be .^ x ^ , ^. e. li times as efficient as 

lb o 

the solid shaft; the angle of torsion or torsional deflection 
will, however, be greater for the hollow shaft. This illustra- 
tion brings out the fact that we can easily see from a considera- 
tion of the stress diagram that the material at the centre of a 
shaft is not used so effectively as that at the outside. This 
result agrees with that which we found for beams (p. 201). 

Numerical Examples on Circular Shafting. — (1) Find 
diameter of wrought-iron shaft to transmit 90 H. P. at 130 revolu- 
tions per minute if the working stress is to be 5000 lbs. per sq. in. 

In this case H.P. = 90, N = 130, and s = 5000 
.'. Working from the general formulae, which are much 
easier to remember than the particular one, we have 



T -- 


5 X TT D3 5000 TT D3 
16 16 


also T = 


90 X 33,000 X 12 
2 TT X 130 


• F)3 - 


16 X 90 X 33,000 x 12 




5000 TT X 2 TT x 130 




= 44*45 ins.^ 


.-. D = 


= ^44-45 




= 3-54 ins. 




Adopt 3J ins. diameter. 



326 THE STRENGTH OF MATERIALS 

(2) A steel shaft 4 in. in diameter is running at 130 revolutions 
per minute and is found to have a twist or " spring " of 9 degrees 
measured upon a length of 30 feet. What horse-power is being 
transmitted, taking G = 12 x 10^ lbs. per sq. in., and what is 
the maximum stress in the shaft ? 

Our general formula is 

js _T_ _ GO 
R l,~~ I 

In our case we are given the following — 

K-Z, 1, _ g^- - -32- -^-^ 

^ = ^° = w '^^^^^' = ^ * 

I = 30 ft. = 360 ins. ; G = 12 x 10^ lbs. per sq. in. 

^ G ^ L 12 X 106 X TT ^ 2 7r2 X 105 . „ 

.'. I = — j—^ = — „„^ ^^ — X Stt = — ^z, in. lbs. 

/ 360 X 20 15 

, ^ ^ H.P. X 33,000 X 12 
but T = 



H.P. = 



27rN 

2 TT X N X 2 TT^ X 105 
33,000 X 12 X 15 

4 TT^ X 130 X 102 



33 X 12 X 15 
= 271 

To test whether the stress is within safe limits we write 
G<9R 



s = 



I 



_ 12 X 10« ^ 

360 ^ 20 ^ 

= 10,500 lbs. per sq. in. nearly. 

(3) What diameter of hollow shaft would you use to transmit 
5000 H.F. at 60 revolutions per minute if the maximum torque 
is IJ times the mean and the safe shear stress is 7,500 lbs. per 
sq. in. Take the internal diameter as half the external. 



TORSION AND TWISTING OF SHAFTS 327 

J ... . H.P. X 33,000 X 12 . „ 
In this case mean torque = ^ — -^^ m. lbs. 

^ 27rN 

5000 X 33,000 X 12 , „ 

= iy ^7,^ m. lbs. 

2 7r X 60 

Tvr ^ + T, 1-5 X 5000 x 33,000 x 12 . ,^ 
Max. torque = T = ^ ^ m.lbs 

T>, . rp _ ^ '^ r' " V 2) J 7,500 TT 15 D3 
^''^^- 16D =T6~'~16~ 

r)3 _ 1;^ X 5000 X 33,000 x 12 x 256 
2 TT X 60 X 7,500 TT X 15 
This gives D = 17'9 inches. 

Adopt 18 inches external diameter . 

* Combined Bending and Torsion. — In a large number 
of cases in practice shafts are subjected to bending as well as 
torsional stresses; a common example occurs in the case of 
a crank shaft and also in the bending stresses caused by the 
weight of the shafts themselves or by pulleys carried between 
the points of support. 

Fig. 145 illustrates the action in an overhung crank shaft 
The driving force p is applied at the point a at the crank pin 
and causes a twisting moment equal to p . a b = p r about the 
axis B c. In the vertical plane we have the couple formed of 
p at A and p at b in an opposite direction ; to preserve 
equilibrium at B an equal and opposite force p must act which 
in the horizontal plane of B c combines with the reactionary 
force p at c. B c may therefore be regarded as a cantilever 
subjected to a bending moment equal to p x b c = p /. It 
is usual to assume the cantilever as extending to the centre 
of the bearing for the purpose of calculating the bending 
moment ; it is difficult to obtain a much more accurate result 
until we know the distribution of the pressures upon the bear- 
ing. Our procedure for the calculation of the combined 
stresses is as follows — 

First find the maximum bending moment M and the 
twisting moment T acting at any point along the length of 



328 



THE STRENGTH OF MATERIALS 



the shaft and calculate the corresponding maximum tensile, 
compressive and shear forces contributed by the bending 
moment and twisting moment respectively. If the shear 
stress contributed from the consideration of the shaft as a 
beam is at all appreciable we should add this stress to the 
maximum shear stress given by the torsion. Let / and s be 
the maximum bending and shear stresses, then, as explained 
on p. 44, we have three alternative formulae to apply, one of 
which gives the resultant shear stress and the other two the 
resultant tensile stress. 




Th( 

(1) 


Fig. 

3y are 
Rankine's formula — 

Equivalent tensile stress 

St. Venant's formula — 
Equivalent tensile stress 


145. 

2^4 






(2) 




4s2\ 




+ ,2 



(3) Guest's formula — 

Equivalent shear stress 



1 + 



= s ^l + 



4^2 



TORSION AND TWISTING OF SHAFTS 329 

Equivalent Bending and Twisting Moments. — It is 
common to express these formulae in terms of equivalent 
bending or twisting moments. 

Now the polar modulus of a solid or hollow circular section 
is twice the ordinary or bending modulus of the same section. 

For the solid circle Z,, = -.,-— and Z = ^^ 

lb 3Z 

.'.we shall have 

_M _ T _ T 
^ ~ Z' ^ ~Z^ ~2Z 
s T 

• • / ~ 2 M 

. • . taking Rankine's formula we have, if Mj, is the equivalent 

B.M. 

M, _ M /^ , /^ ^ T2^ 

IP. 



Equivalent tensile stress = ^^ = -^ (l +^/l + 



.■.M. = |(l + ^1+|;) (4a) 



'' 2 



|(m + VM^ + T^) (46) 



St. Venant's formula would give 

or = g- + g VW^pT^ (56) 

• Guest's formula will give 

T M /~ T2 

Equivalent shear stress = -^ = ^-^ /i j_ 

M / T2 



or = VMM^T2 (66) 

For ductile materials, such as steel, formulae (6) are recom- 
mended for use in design, the safe shear stress being used for 



330 THE STRENGTH OF MATERIALS 

determining the necessary value of the polar modulus Z^„ 
to carry the twisting moment. 

For brittle materials, such as cast iron, formulae (4) or (5) 
should be used, the St. Venant formula being recommended 
as the more reliable of the two ; the safe tensile stress should 
be taken in this case. 

There has been considerable confusion with these formulae 
because Rankine's formula is often given as an equivalent 
twisting moment and has been compared with Guest's formula 
for the same working stress; the point to keep in mind is 
that if Rankine's formula is used the safe tensile stress and 
bending modulus should be taken, but if Guest's formula is 
adopted the safe shear stress and polar modulus shoukl be 
used. 

Numerical Example. — What must be the diameter of a 
solid shaft to transmit a twisting moment of 160 ft. tons and a 
bending moment of 40 ft. tons, the tensile stress being limited 
to 4 tons per sq. in. ? What diameter would you use if the shear 
stress is limited to 3 tons per sq. in. 



Let D be the diameter of the shaft. 




Then 


s 


T 

~7rD3 

16 


16 T 








/ 


IVI 

7rD3 


32 M 








-U' 


32 


4 '^ 4 






3 
• • 4 


4s^ 

+ /2 ^ 


^/l + 


4 X (1()T)2 
(32M)2 




Vi + 






^1 + 


\40/ 








= •75 + 


4VI7 


= 5-9 



TORSION AND TWISTING OF SHAFTS 331 

,5-9/ 5-9 32 X 40 X 12 

. • . Max. stress = 4 = —7,-^ = -^^ x tv^ 

2 2 ttD'* 

-r.„ 5-9 X 32 X 40 X 12 ^ ^.„ , 
D^ = 5 = 3,600 nearly, 

O TT 

D — 15*3 inches about. 

/160\^2 



Equivalent shear stress = <5 ( a/ 



1+V40 



_ 16 X 40 X 12 X 4-12 
16 X 40 X 12 X 4-12 



D3 = 



37r 

D = 15 inches about. 

Shafts with Axial Pull or Thrust. — If in adition to the 
torsional stresses (and bending stresses if they occur) there 

P Q 

exists an axial pull (P) or thrust Q we add -. or . to the bend- 
ing stress to get the value of / to use in the formulae. In the 
case of end thrust acting, when the direct stress is taken as 
a criterion, the resultant direct stress should not exceed the 
safe stress upon the shaft considered as a column, and the 
design should be treated similarly to that of a column with an 
eccentric load (p. 301). In the case of steel shafts, which are 
most common, we suggest that a double test should be applied ; 
the shaft should be designed to carry the equivalent direct 
stress as a compressive stress on a column and also by the 
Guest formula, the diameter chosen being the greater of the 
two results. 

Torsional Resilience. — As we have already explained in 
connection with resilience in bending, the work done bj^ a 
couple is equal to the product of the couple into the angle 
turned through. 

If, therefore, the angle turned through is 0, the work done 

T6 
by the couple which increases gradually from to T is -^ • 

R" I ~l, 



332 THE STRENGTH OF MATERIALS 

. • . Work stored in shaft = ^ = c^' i^ • ?>. -r> = ~ '' ta 

2 

21, .sH 



(1 



G D2 

For a solid shaft -r^'l = , ^ 
D- lb 

the volume of the shaft = —j— .1 = Y 

4 

. 2jy_v 

• • D2 4 

i.e. Work stored in solid shaft = -7 ^^ • V 

4G 

52 

. • . Resilience = work stored in unit volume = -r-^ .... (2) 

4G ^ ' 

2 
If we take G = E (cf . p. 12) this gives 

o 

5^2 

Resilience = -^^^- (3) 

For a hollow shaft of external diameter D and internal 
diameter D^ 

2I,Z _ TT (D4 - D,4) . I _ ^ (D2 4- D^2) ( D2 _ 1) ^2) Z 
D2 ~ 16 D2 ~ 16 D2 

and V = '^(5i^-»A^ 
4 

2 I, Z /D2 + Di2\ V 



• • D2 ^ p2 y • 4 

.-. Resilience = ^^^ |l + (^J^^^ (4) 

In the limiting case of a very thin tube j} approaches 1 
and we have 

^2 

Resilience = -^-p (5) 

Torsion of Non-Circular Shafts.— As we have akeady 
indicated on p. 320, the ordinary theory of torsion is true only 
for circular sections because in other sections the sections 



TORSION AND TWISTING OF SHAFTS 333 



originally plane become bent out of the plane upon twisting. 
The following cases have been worked out fully by St. 
Venant, who has giVen the following approximate formulae — 



Section. 



Relation of Maximum 

Stress to Twisting 

Moment. 




T 




T = 5 . -208 S3 



T~ 

H 



B 



Any symmetrical sec- 
tion not containing re- 
entrant angles. 



T = 



sBH2 



3 + 1- 



H 
B 



T 



5 A* 



40 \, . y 



Angles of Torsion in 
Degrees. 



_292TZ((^2^D2) 



GD3# 



410 TZ 



205 TJ (B2+H2) 
G B3"H3 



40I,,TZ 
A^G 



Where A = area of section 

ly = j)olar moment of Inertia 
y = distance of farthest edge 
from centre of shaft 



Numerical Example. — A square steel shaft is required for 
transmitting power to a SO-ton overhead travelling crane. The load 
is lifted at a rate of 40 ft. per minute. Taking the mechanical 



334 THE STRENGTH OF IMATERIALS 

efficiency of the crane gearing as 35 %, calculate the necessary 
size of shaft to run at 160 revolutions per minute. The twist 
must not exceed 1° in a length equal to 30 times the side of the 
square. Take G = 13 x 10^ lbs. per sq. in. 

Work per minute to lift weight = 30 x 40 ft. tons. 
If rj of crane = 35 %. 

w w V. r 1 • . 30 X 40 X 100 
Work to be supplied per minute = ^ 

.'.If revolutions per minute = 160 

r^ 30 X 40 X 100 ,. , ^ 

Torque = ,^ ^ rrrAft. tons = T 

^ 3t) X 2 TT X 160 



I = 



3-42 ft. tons. 

410 X 3-42 X 2,240 x 12 x 30 S 
13 X 10« . S* 



^.3 _ 410 X 3-42 X 2,240 x 12 x 30 
13 X 10« 

S = 4J inches (sa}^). 




Fig. 146. 

Effect of Keyways upon the Torsion of a Circular 
Shaft. — Professor H. F. ^Nloore of Illinois University has 
given the following results of experimental investigations 
upon shafts with kej'wa}^ grooves cut in them. 

Strength of cut shaft _ '26 lid 

Strength of uncut shaft D D 

Angle of torsion of cut shaft _ , * 4 6 , '7 d 

Angle of torsion of uncut shaft D D 



TORSION AND TWISTING OF SHAFTS 



335 



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CHAPTER XII 

SPRINGS 

Springs may be regarded as devices for storing up energy 
in the form of resilience and are used either as a storage for 
energy, as in clocks, phonographs, etc., or else are used in 
order to absorb excess energy which would otherwise do 
damage, buffer and carriage springs being very common 
examples. 

The best form of spring will be that which will absorb the 
greatest amount of energy for a given stress, and in the case 
of springs placed upon trains it should be remembered that 
the kinetic energy of the spring itself will add to the energy 
that has to be absorbed. Failure to appreciate this fact has 
led many people to suggest that railway collisions could be 
prevented by the use of heavy springs placed in front, whereas 
the weight of spring necessary to do this would be so great 
that its advantage would be lost. 

Since all elastic bodies have resilience, all forms such as 
ties, struts, beams and shafts are strictly springs, but in ordi- 
nary form they are too stiff to be of use. The j)rincipal 
forms of springs may be divided into torsion springs and 
bending springs. 

We can consider the relative values of tensile, bending and 
torsion springs by comparing the relative resiliences. For a 
beam of uniform strength loaded at the centre such as occurs 
in a leaf spring we have 

/2 

Resilience = ir^ 
6E 



SPRINGS 337 

For tension we have 

Resilience = ^-^ 
2 E 

For torsion we have 

Resilience = -.-^ = ^-^7 
4 (jT o hi 

4 

Taking 5 = ^ / this would give 

Resilience = ^^-4^ 
5E 

or, on the Guest theory, if <s = ■^ we shall have 
Resilience = 



32 E 

The pure tension spring, therefore, which is practically 
never used in practice, is the most economical, and the relative 
economy of the torsion and bending springs depends upon 
the view taken as to the relative safe stresses for the two 
cases. 

Time of Vibration of a Spring. — The vibration of a 
spring follows the laws of simple harmonic motion, so that 
the following general formula will enable the time of vibra- 
tion to be obtained. 

Time of complete vibration 

= ^ _ 2 TT / Weight of sprin g ^ _ ^ ^^^ 

y g X Force to cause unit deflection 

. • . number of complete vibrations per second == - 

z 

TORSION SPRINGS 

Close-Coiled Helical Springs. — If a helical spring is so 
closely coiled that each turn is practically a plane, the stresses 
upon the material will be almost pure torsion. The twisting 
moment. Fig. 147, will be W R and the spring will be equiva- 
lent to a shaft of diameter d and of length I equal to the total 
length of wire in the spring, ^. e. if n are the number of turns, 
I =. 2 TT R ?^ (approx.) ; the torque applied to this equivalent 
shaft will be W R as indicated. 



338 



THE STRENGTH OF ]\L\TERIALS 



By our general torsion formnla we have 

s _T _Gf 

2 

/T nvR 

•*• ^ " GI,, " Gjd^ 
32 
32nVR 
~ GVci^ 



(2) 




LUILUIL 



turns 



Jlllli 



in 



L = 27rR 



^^ 



d 





Fig. 147. — Close-coiled Helical Springs. 



Now consider a very short length a of the spring and 
suppose that 6 is the angle of torsion of this short length; 
then due to this angle the weight W will go down by an 
amount R ^ as indicated in Fig. 147. This is true for every 



SPRINGS 339 

short length and the separate deflections due to each piece 
add together giving a total deflection 

. ^ , 32WI12; .„ . 

=^"G7y (^^^ 

If n is the number of turns we have I = 2 -n-Jln 

_ 64jT Kn W R2 

= Gd^ "^ (^^) 

-Gd^ -^ (*^> 

. • . the load W to cause a deflection 8 is given by 

W=7rf|3-.8 (5a) 

64 R^ ?i ^ 

Grl^ 

= 8T5^-^ (3^) 

We might have obtained our result as follows from the con- 
sideration of the torsional resilience (p. 331) — 

Torsional resilience = -. ^ 

4G 

. • . Work absorbed = ^-p x volume = Work done by weight 

_ W8 

2 

. _ 16 W R ^ _ Trd^l 

^ ~ Tvd^ ' ^ ~ 4 

W8 _ 162 w^ R2 . TT (^2 / 
•'•"2 " TT^d^ 4G.4 
_ 16 W^ R^ I 

~ TT # G 

, 32WR2? 

6 = 7>ivm— as beiore. 

Time of Vibration. — Putting 8 = 1 in equation (5) we 
get the force to cause unit deflection ; this can then be put 
into equation (1) to find the time of swing, remembering that 
if all the other dimensions are in inch units, g (the gravity 
acceleration) must be also reduced to inch units. 



340 THE STRENGTH OF MATERIALS 

Numerical Examples on Closely Coiled Springs. — (1) 
A closely wound helical spring is formed of 30 coils of |- in. 
round wire, the mean diameter of the coil being 4 ins. What 
axial load will produce a shear stress of 9 tons per sq. in., and 
if G is 4,700 tons per sq. in., wlmt will he the extension of the 
spring under the load ? 

s y ird^ 



T = Wx2= ,, 

1(3 



1\3 



^^^ ~ 16 

9 TT 

W = Wo a A tons 

32 X 64 



lbs. 



_ 9 7r X 2,240 
" 32 X 64 
= 30-9 lbs. 

_ 64 \V W n 
^ = ' Gd^ 

64 X 30-9 X 30 X 8 X 256 , , _ . 

= 4,700-x 2,240 ^ \}21L2111 

Before this extension occurred the spring would have 
ceased to be closely wound if the load were such as to stretch 
the spring, and the spring would have been fully closed if 
the load were such as to compress the spring. 

(2) // a closely wound helical spring made of wire J in. in 
diameter has 10 coils, each 4 ins. mean diameter, find the fre- 
quency of the free vibrations when it carries a load of 15 lbs. 
{taking G = 12 x 10® lbs. per sq. in.). 

\ ^ X Force to cause unit displacement 
Putting 8 = 1 in equation {oh) we have 

Force to cause unit displacement =jjpv3 

(i)* X 12 X 106 Qi^^i, 
8 X 43 X 10 



Frequency = 



/ — 9 - ^ /, seconds 

" \ 32 X 12 X 9-156 

1 _ 1 / 32 x~12 X 9156 

t 27r\ "15 

= 2 43 per second. 



SPRINGS 



341 



It should be noted that in the above calculation we have 
neglected the weight of the spring itself, so that the result 
can only be regarded as approximate. 

Weight of Springs. — A cubic inch of steel weighs about 
•284 lb. 



= -284 X 



.'. Weight of spring = "284 x volume 

7V_dH 

4 

TV d? irT> n 
4 



= -284 X 

= '^ d'^T> n very nearly. 



Safe Loads on Circular Springs. — The following are 
the highest safe torsion stresses upon steel spring wire found 
by experiments by Mr. Wilson Hartnell — 



Diameter of Wire. 
1 

2 


Safe stress in lbs. 
per sq. inch. 


Safe Load on Spring 
(lbs.) 


70,000 
60,000 
50,000 


429 
D 

1,240 
D 

2,450 
D 



D = mean diameter of spring in inches. 



we have W 



From the formula T 
2 X '\^Qs(P 



WD 

2 



D 



from which the values in the 



third column of the table are obtained. 

Restriction on Use of FoRMUL.g]:. — When ine springs 
are used so that when stressed they are shorter than when 
unloaded {i. e. if the load were to act upwards in Fig. 147), 
it should be remembered that the spring may become shut 
before the safe load has been reached ; this does not diminish 
its strength, but impairs its value as a spring. 



342 THE STRENGTH OF IVIATERIALS 

Alignment Charts for Close-coiled Circular Springs. 



DIAMETER OF SPRING 



\ «A ■■ ■ ll ■ I I I' M l I . I . I1 . .I I ■ ■ ^ - I ■ I I t ■ I 



►. — -; ;C 00 -o 



/ \ 

ALLOWABLE WORKING STRESS 



5» > 



Sj5 / 

Jo / 

?" / 

</> / 



r o p o c S 5 5 — 



i I i i i 



DIAMETER OF WIRE 
l -" l- ■■■ I ' ' ' / ' ' ' L ' — T 



/ 

I 

I 

t 

I 



LOAD ON SPRING POUNDS 

§ I MIII .1 I . Lhl 'l ■ I I || m/ ||[|| | I I . I h ll. M I . ■ |.|i|i I ' I ' |l|.'|ii . 'l ■ 

5^0 0000000 o o 00 — CO <" 

J-O COOQOOOO O w 



Fig. 148. 



— Figs. 148, 149 show alignment charts * for the strength and 

* For an explanation of the principle of these charts see a booklet 
by the author on Alignment Charts, published by Messrs. Chapman 
and Hall, Ltd. [Price 1/3 net.] 



SPRINGS 



343 



deflection of round wire helical springs, taken from an article 
by Mr. F. Fitchett in Machinery for January 14, 1915. 



o ooooooo 

P ' 1 1 r I I I I I 1 1 1 1 I I I 1 1 1 1 1 1 1 1 1 I ' I 



X. 



§$gg s ss s g 



l " ' >' I 



LOAD ON SPRING. POUNDS 



DEFLECTION pF SPRIN9. INCHES 



.|... I ■ t ,\.\.i. V 



t ■ I ■ I.I .1.1 



^ V« 9> O QC tC 



w ♦ wo>^«<cr: 






I I I I I ' I ' I — ' — v 



if ^* 



DIAMETER QF WIRE 
I / 






/ I 
DIAMETER OF SPRING 



±T- 



COEFFICIENT OF / 
RIGIDITY ' 

I ■ I ' I ■ t ■! ■ I 



I . I I t i / l I I I I II iIi m I I ) 



ro u u '^ u< 9k 



| " 't ■ t ■ t ■ I'l'l'l ' M 
a> ^ OB « — — — —~7i'^ 



o — >o ■■^■^i^en 



ai 00 



P !^ ■'^ -* / 



r i'i'i ' " I """ 

"- '^ oc *J ^ tn ^ 

oo o o o o o 



NUMBER OF COILS 
iii|ii I I W iiiii I I 1 I I I 1 t I 



-r (C off ^J * (ji ^ 



I I 



FiG.^ 149. 



To use the charts we proceed as follows : Suppose, for 
instance, mean diameter of coil = 3 in., diameter wire = f in., 



344 THE STRENGTH OF MATERIALS 

safe stress = 60,000 lbs. per sq. in. On Fig. 148 connect 
" diameter of spring " to " safe working stress '' and note 
the intersection on the dotted axis. Connect this point of 
axis through " diameter of wire " on to " safe load," which 
will give the result required, viz. 415 lbs. approximately. 
The chart can be used similarly to find anj^ one of the four 
quantities if the other three are given. 

To use the deflection chart of Fig. 149 take, for example, 
a load of 16 lbs., J in. diameter wire, 2i in. diameter of spring 
consisting of 20 coils, using G = 10,000,000 lbs. per sq. in. 
Connect " number of coils " to " coefficient of rigidity '' and 
note intersection on the right-hand intersection or " support 
line. Then join " diameter of wire " to " diameter of spring " 
and note intersection on centre support. Join these two 
intersections to meet the right-hand support and connect the 
point thus obtained to " load on spring '' and produce to the 
'' deflection " which gives 1 inch approx. 

Springs of Square Section Wire. — If our spring is like 
that shown in Fig. 147 with the exception that the wire is 
square instead of round in section, the length of each side 
being S, Ave can proceed as follows — 

410 T / 
From p. 333 = r^(^^ (degrees) 

711T/ , ,. - 

= G S* (^^^^1^^"^) 

andT = s -208 S3 

WD 

Now T = W R = ^ 

„, s X -208 S3 -416 5 S3 .,. 

••• ^^ = D" =-^- ^^^ 

2 
711R X WR; 

G S* 

GS^ 
1-78 WD2/ 

"GS* •• ^^^ 



8 = R^ = 



puttii 


ig I 

8 


SPRINGS 
r787rWD3 


n 




GS* 
5-6WD3 7i 

GS* 





345 



(3) 



Taking the same safe stresses as for round wire we get 
from formula (1) the following formulae for the safe loads 
on square wire springs — 



Side of Square (in.) 


Safe Load in lbs. 




455 
D 

1316 
D 

2660 
I) 



Comparison of Square Section and Circular Section 
Springs. — The volume of the square spring = S^ Z = V 



T B 
. Work absorbed = -^ 



208 s S 3 7;^! IT I 

GS*"^^" 

7-11 X -208 5 S3 Z 



X 



2 

•208_5_S3 
2"~" 

G" 
154 ^2 



GS^ 



154 



Resilience = - —^ — 



G 

154 <§2 
G 



This is less than for a solid circular spring, the ratio of 

25 

circular to square resilience being 7 — ^ = 1 • 62 . Also for a spring 

of given diameter the load carried is greater for a circular 
section than for a square section of the same area. 



d 




T. 


1128 


• t; 


4 X -208 



346 THE STRENGTH OF ^L\TERIALS 

For the circular section we have 

rj. s X ird^ 8 d 

for square T = 208 s S^ = 208 5 S x area 
T _ d 

■'• T. ~ 4 X -208 8 

K — - = S-, /. e. if the areas are equal as suggested 
4 

1128 S 
= 1-36. 



Weight for iveight, therefore, a closely coiled circular section 
helical spring of given diameter is 136 times as strong and will 
absorb 162 tirnes as much energy as one of square section of 
the same diameter. 

Open-Coiled Helical Springs. — In the case of open- 
coiled helical springs the stress is principally a torsioned one, 
so that we wiU deal with it here although there is also bending 
stress. 

Referring to Fig. 150. let the centre line of the spring at 
any point be inclined at an angle a to the horizontal ; then 
the normal section plane x x of the wire will be at an angle 
a to the vertical load W. The load W has a moment W R 
about the line centre o of the section and this moment has 
a component o « = W R cos a which is a moment in the plane 
X X and causes a t^visting action, and a component a 6 = W R 
sin a which is a moment normal to the plane x x and causes 
a bending action. 

.-. T = WR cosal 

M = WRsina/ 

It is not altogether easy to follow this resolution into a twist- 
ing and a bending moment at first, parth* because it is not 
very easy to give a very clear diagram with the forces acting 
in different planes, but a Little consideration of the problem 
will probably remove the difficulty. 



SPRINGS 



347 



The angle of torsion in the plane x x will be given by 
.'. Work done against torsional stress = -^ = 901 

W2 R2 C0S2 a I ' 

- ~^GX ^ ^ 




Fig. 150. — Open-coiled Springs. 
The bending moment is constant, so that (see p. 264) 



Work done against bending stress 

_ W2 R2 sin2 a I 

2EI 



2EI 



(3) 




348 THE STRENGTH OF MATERIALS 

. • . Total work done against stress 

W2 R2 I /cos2 a sin2 a\ 

= 2 Ul, + ElJ (^) 

But total work done against stress = Work done by 

weight = 2 

•■■^-^^^^Kov + ti") (^) 

Taking a solid circular section we have 2 I = I^, = 
. 32 W R2 Z /cos2 a , 2 sin2 a' 

_ 32 W R2 / 

(and taking ^ = 2*5 

32WR2Z 2 ^ Q • 2 ^ .AN 

= — /T' -74 (COS^ a + '8 Sni^ a) (6) 

= ri ^4 (COS'^ a + '8 sni- a) 

= G.# <1- 2'™'") (') 

The movement due to tAvisting per unit length of the coil 
will be equal to R 6.,, where (9,, is the angle of torsion, and takes 
place in the -plsme x x. This is equivalent to a movement 
d e vertically and a horizontal movement e /. 

.-. S,, = R^,, cos a . ..(8) 

^ / = R 0, sin a (9) 

This movement e f tends to increase the number of the 
turns of wire. 

The bending deflection upon a unit length will be equal to 
R ^B, where 6'„ is the angular change in the centre line of the 
beam. This movement takes place in the plane y y and will 
tend to unwind the coil ; it has a vertical component g h, 
which is the part of the deflection contributed by the bend- 



SPRINGS 349 

ing and a horizontal component h j which is an unwinding 
tendency and opposes the winding-up strain ef. 

.' . 8„ = R, 0,^ . sin a 
h j = — R (9„ cos a 

T 

Now 0^ = ^ J- per unit length 

M 

^„ = =^j per unit length because the bending moment is 

constant.* 

. • . per unit length 

T n X- o , cs R T cos a , RMsina 
denection = o^ + \ = ^ t H ^cTt — 

_ W^R2 cos2^ W R2 sin^ a 
~ ~" GI,, ^ EI 

The deflection will be the same for each unit of length — 

This agrees with our equation (5) obtained in a different 
manner. 

Angular winding-up movement per unit length 
■ _ e / — hj 
~ R 
= 6^ sin a — ^,5 cos a 

,^_ _, , sin a COS a sin a cos a 
W R 



GI„ EI 

Total winding-up movement 

= /3 = W R ? sin a cos «■ f p ^ ~ -^ -r 



32 W R ? sin a cos a /-^ _ 2 G\ ,^^. 



7r#G V E 

for a solid circular section 
_ 3 2 WR Z sin a cos a G _ 2 
~ 5 TT # G" ^^ E ~ 5 

* See p. 249, and note in Fig. 121 that 9 = ^ = 4 ^f C C is unity 

.-. e= 1- ^. 



350 



THE .STRENGTH OF MATEKIALJS 



This is a maximum for a = 45"^, 

If }) is the i)itch of the coil and n is the number of turns 
/ __ n^/ p'- - -MV^ 

Comparison of Close-coiled and Open-Coiled 

Springs. — Comparing result (7) with ecj^uation (36), p. 33*J, 

for the close-coiled spring, we have 

8 for open-coiled spring , , ^ • o x 

. .. , ^ . ^ ^ (1 — 2 sm- a) = 



8 for close-coiled spring 



in 



roi 




^^^ 














98 






^ 


^. 


















N 


N 










96 










\ 


\ 


















\ 








"OJ 














\ 






II 














\ 


\ 




c 

JO 

"a 

B 














' \- 














\ 


o 

2 
S 






: \ 



2.0 



30 



"W 



O JO 

e (degrees) 

Fig. 151. — Correcting Coefificients for Open-coiled Springs. 

m may then be regarded as a correction coefficient, values of 
which are given in Fig. 151 for various values of a. 

Stresses in Wire. — The stresses in the wire can be found 
by calculating the separate bending and shear stresses and 
combining them in the manner described to find the equiva- 
lent simple direct or shear stress. 

The twisting and bending strains both cause a tendency 



SPRINGS 351 

for the free end of the coil to turn about the vertical axis v v, 
thus altering the effective number of coils. 

BENDING SPRINGS 

Leaf or Plate Springs. — If we consider a leaf or plate 
spring of the type shown in Fig. 153 and note that the plates 
are bent to the same radius so that they contact only at their 
edges, we see that each plate may be regarded as supported 
at its point of contact with the one below it, the load trans- 

W 

mitted at its overhanging end being -^ . (See also Fig. 152.) 

The B.M. diagram for each plate comes therefore as shown. 

In order that the spring may close practically fiat, the 

curvature of each plate must remain constant after bending, 

i.e. the radius of each plate after bending must be the same. 

1 M 
But from p. 249 ^ = ^^ 
^ R EI 

M . 

. • . Since E is constant ^ is constant. 

Between b and b', the B.M. is constant so that the section 
is constant, but for A B and a' b' the B.M. varies in the 
triangular manner shown, so that the section must vary 

so as to keep ^ constant. 

This can be done by making the ends triangular in plan, 
the thickness being constant. 

Then at any point at distance x from a 

12 

M = ^\^ 

M _ 6 6 X rf3 
•'• I ~ Wa: 

and ^ = 7, by similar As 

M 6c?3 

. • . T =1X7 7/ = constant 
I WZ 



352 



THE STRENGTH OF MATERIALS 




Fig. 152. — Stresses in Plate Springs. 

Another way would be to keep the ends square and to vary 
the thickness as indicated. 

Then I ^ *jf , M = 'Y 



M 



For -p to be constant d/ 



SPRINGS 

d^ X X 



353 



and if the lap were 



I ^^ — "'•• ~ r 

contoured so that the relation held, the necessary conditions 
would be satisfied. 

Now suppose that there are n plates and that all but the 
top one are cut longitudinally through the centre and placed 
as shown in Fig. 152 they would make up the diamond-shaped 




Fig. 153. — Plate Springs. 

figure shown. The deflection, i. e. the upward vertical 
movement from its initial curved position, for such a single 
jDlate, which bends to a circular one will be very nearly equal 

MP 



to 



8EI 



Now I 



?i b d^ 
"12" 



3MJ2 

~^2Ew6# 



A A 



354 THE STRENGTH OF MATERIALS 

For a centrally loaded beam M = —^ 

' ' SEnbd^ ^^ 

In a test of such a spring it will be found that the friction 
between the plates will cause the deflection to be less than this 
with an increasing load and to be more as the load is reduced. 
It is common to test such springs by loading them until 
the plate is flat; we then have 8 = 8,„ and we get from 
equation (1) the following value for the test or proof load W^ 

Stress in Plates. — The stress in the plates will be con- 
stant along their length because their depth as well as their 
moment of inertia is constant. 

•••/ = 



M d 

I ^ 2 

Wl 


d 


. nb d^ 
^' 12 
3W? 


' 2 



~ 2nbd'' ^^^ 

Derivation of Deflection from Resilience. — The 
formula for deflection may be derived from the resilience as 
follows — 

42 

Resilience (see p. 273) = J^ 

.*. Total work done in stressing = y!^ x vol. 

- ^-- Id — 
"" 6E •^'^- 2 

JWS _ f^lndb ^ 

•*• 2 ~ 12 E 

dWU Hn'^db 

~ In^b^ d^ xl2E 

^ 3W ^ P 

~ I6nbd^ . E 

.*. d = c,T->- 1. Ti as beiore. 



SPRINGS 



355 



Numerical Example. — A lamiyiated plate spring of 40 
inches span has 12 plates, each -375 inch thick and 3*40 inches 
wide. Calculate the deflection when carrying a central load of 
4 tons, taking E = 11,600 tons per sq. in. 

By formula (2) we have 

3WP 



8 = 



S^nbd^ 

3 X 4 X 40 X 40 X 40 



8 X 11,600 X 12 X 3-40 x -3753 
= 3-84 inches. 




Fig. 154. — Piston Rings. 

Piston Rings. — Springs in the form of split-rings are 
placed around pistons in oil, gas and steam engines to prevent 
escape of the working fluid past the piston, and such rings 
should be designed so as to give as constant a pressure as 
possible all round the cylinder. The necessary variation in 
thickness has been investigated by Professor Robinson in 
the following manner. 

Let T>, Fig. 154, be the point of maximum thickness to 
at the centre of the ring which was initially circular on the 
outside and is sprung into position so that it is still circular 
on the outside. 

Consider a length a B of the ring, the thickness of the 
ring at the point b being te and the breadth throughout 



356 



THE STPxEXGTH OF MATERIALS 



being b. ^ is the angle which the arc b D subtends at the 
centre. 

Let R be the radius of the spring when bent and let R„ 
be the radius at b when in the unstrained condition indicated 
in dotted lines. 

If }) is the pressure per sq. in. we have 

B 
p = 7^ . A B (chord) . b = 2 pb R cos .^ (1) 

and the bending moment at p is equal to 




Fig. 155. — Piston Rings. 

By a modification of Fig. 121, p. 249, assuming a small 
initial radius of curvature R„ we shall get 

\R lij EI 
as a first approximation. 



I _ 1 
R R, 



EI 



2 pb Fx.^ cos^-^ 



E 



bte^ 
12 



24 p . R2 cos2 2 



Ete' 



(3) 



SPRINGS 357 

At the point d, ^ = and te = t^ 

]. _ J. _ 24y R2 
• • R R„ ~ E ^ 3 (*) 

Now R„ and R have to be the same in each position 

24 25R2cos2|- .. _^. 
^ 2 _ 24 2? R^ 

•'• El? ~ Ee 

If, therefore, the pressure 'p is to be constant 

te^ 2 ^ 

1} = ^^^ 2 

•••| == V'°''l ^^^ 

Fig. 156 shows vahies of te in terms of f^ for the various 
values of 6. 

Neglecting the additional stress due to the curvature of 
the bar (see ChajD. XIX.) we have at the point d 

^ I ^VR R„ 

2 

E L / 1 1 



•'• ^ 2 \R R, 

112/ 

^•^•r-r;=^e7: (') 

Putting this result in (4) we get 

I = l2pf-^ (7) 

.-. ^. = rVt^ <^' 

To find the necessary initial radius we have 

i- = i - ^ . (9) 

R„ R E^, ^^^ 

Numerical Example. — Taking E == 16 x 10^ lbs. per sq. 
in. and the working stress 4000 Ihs. per sq. in., find the necessary 
thickness and original external diameter for a cast-iron piston 
ring for a cylinder 20 inches in diameter, the necessary pressure 
being 3 lbs. per sq. in. 



358 THE STRENGTH OF MATERIALS 

From equation (8) 
^„ = 10 X 



12 X 3 



4000 



= "95 in. nearl3^ 



This is less than is usually used in practice. 



/so 



/60 



140 



(20 



IOC 



3g 



GO 



"KP 






20 



. . . 

. V- 1 



•2 



o 



lo '8 O '^ 

Values of f^ — t^^. 

Fig. 156. — Thickness of Piston Rings for Uniforni Pressure. 

Unwin and Mellanby * give total depth of all packing 
ngs 

* Elements of Machine Design (Longmans), Part II (1912). 



SPRINGS 359 



= , ^ + '6 in. for steam engines 
15 

= -f-- + 2*4 ins. for gas and oil engines 

= ^7^ for petrol engines. 
Taking, therefore, a steam engine with rings we should have 

= -64 in. 
From equation (9) 



11 2 X 4000 



• ■ R, R 16 X 10« X -95 

•1900 
= -09947 
R„ = 10053 
.*. original diameter = 20*11 nearly. 

Ring" of Uniform Thickness. — If the ring is of the same 
thickness t throughout, R will be constant, but R^ should vary 
in accordance with the following treatment — 

We have as before in equation (3) 

, , 24 ^ R2 cos^ ^ 

R ~ K "" El3 

, , 24 2? R^ cos^ -^ 
•'•R, ^ R El3 

■ •■K^ ^''^ , (10) 

E ^3 _ 24 2? r2 1- 

For given values of t and f, R^ can be found by this formula 
for different angular positions, and it will be found that the 
curve for the initial shape of the ring differs considerably 
from a circle, so that rings made by cutting out parts from 
circular rings and springing into position will not give a 
uniform pressure. 



360 



THE STREXGTH OF MATERIALS 



* Plane Spiral Springs. — Consider a short length a b 
(Fig. 157) of a plane spiral spring, the free end of which is 
pulled with a force W. 

Then the bending moment acting on this short length 




Fig. 157. — Flat Spiral Springs. 

. • . If 8 ^ is the change in angle between the tangents at 
the two ends 

^0 ==^ (seep. 250) 

Ss 



W.r X 



EI 



Total chano;e in ansle 



= ^-2 



M 3 5 W 



EI EI 

if E and I are constant as is usual. 



.r 8 s 



SPRINGS 361 

% X S s = 1st moment of spring constant about x x 
= length of spring x r (approx.) 
= Ir 
. Wlr 



EI 



(1) 



If the spring is wound up to produce a tensile force W at 
the end, the torque T which must be applied to the shaft 
will be equal to W r. 

Also 6 will be the angle turned through by the shaft, so 
that work stored up in spring 



2 E I ^^ 

If the breadth of the spring is b and its thickness t, 
_ bj^ 
6 

6M 6M 



. • . Bending stress = / = 



bt' 



The maximum B.M. occurs at the point and is approxi- 
mately equal to 2 Wr. 

12 Wr 

• 12 r 



Putting this result in (2) we have 

144 r2 . ^ 
/2 bH"^ .1 



/2 J2 li I 

Work stored = j-i .— «— fr^rr 
144 r^ . 2 E I 



b t^ 
288 E . ^- 



24 E 

/2 

Resilience = ^. „ 
24 E 



Pbtl 

X volume 



24 E 



362 THE STRENGTH OF MATERIALS 

This is relatively small because the material is not used very 
economicalh^ parts of the spring being much more highly 
stressed than others. 

Close-coiled Helical Springs under Bending Stress. 

— If instead of subjecting a close-coiled helical spring to an 
axial load we subject it to a twisting action tending to unwind 
or wind up the spring, the whole spring will be subjected to 
a bending moment equal to the torque T applied. 

Therefore, if we neglect the effect of the curvature of the 
wire upon the stresses in it, we shall have, if 6 is the angle 
by which the spring winds up or unwinds — 

T f) T^ I 
Work stored = -^ = o^Fl" ^^^^ ^' ^^"^^ 

• • ^ - E"i 

Circular Section. — For a round wire of diameter d, 

d^ 



I = 



64 



and / • -oq" = 1 



i.e./ 



32 T 



Work stored = 



Resilience = 



77 d^ 



32 . 32 . 2 E Trd^ 
P .T-d^l _ P X volume 
32 E ■ ~ 8E 



8E 

Tl 64 T Z 



radians 



, " ~ E I 7T d^.E 

1,168 TZ , 
— — I decrees. 

Rectangular Section.— If the depth of the section is b 

b t^ 
and the thickness is f , I = -, ^^ 

and T = Lipl 



SPRINGS 



363 



Work stored = 



12 TZ 

E673 

688T^Z 
E h t^^ 



radians 



degrees. 



Resilience = 



2EI 




/2 62 ^4 


.1 


2 X 36. 


E.bt^ 


fnti 


f 


6E ~ 


6E 


P 





12 



X volume 



6E 



SuMMAHY OE Resilience oe Vauious Types of Spring. 



Type of Spring. 



Pure tension 

Pure torsion on close-coiled helical spring (circular shaft) 

„ „ „ „ „ (square shaft) 
Plate spring 



Plane spiral spring 



Close-coiled helical springs with twisting action causing 
bending stresses — • 



Circular section . 
Rectangular section. 



Resilience. 



2E 

4G 

■154 s^ 

~G ■ 

P 
6E 

_£_ 
24 E 



8E 

P 
6E 



CHAPTER XIII 

THE TESTING OF MATERIALS 

Testing Machines. — In most types of testing machines 
the loads are applied through a system of levers and are so 
arranged that the levers are connected to one end of the 
specimen (or in the case of bending tests to the supports), and 
that a force is exerted by an hydraulic ram or screw gear to 
the other end, the lever system " floating " when the force 
exerted is equal to that applied to the levers. In this way 
additional weights can be put on to the levers without causing 
a shock in the specimen, because such additional weight does 
not come on to the specimen until the hydraulic ram or screw 
gear is operated further. We will describe some of the most 
common types of testing machines. 

WICKSTEED-BUCKTON SiNGLE LeVER VERTICAL TESTING 

Machine. — This type of testing machine, a photograph of 
which is shown in Fig. 158, was designed by Mr. J. H. Wick- 
steed, and is manufactured by Messrs. Joshua Buckton & Co., 
of Leeds. The form shown in the photograph is belt driven, 
the power being transmitted by toothed gearing to the screw 
at the base of the machine, but hydraulic rams are commonly 
employed to exert the necessary test force. This particular 
machine has a capacity of 30 tons, machines of this tj-pe being 
obtainable for capacities ranging from 5 to 100 tons, and can 
be employed for tests in tension, compression, bending, shear 
and torsion. 

Fig. 159 shows diagrammaticaUy the action of the machine, 

an hydraulic ram drive being shown. 

364 




fcJD 






> 

o 



ffl 



00 






THE TESTING OF MATERIALS 



365 



A horizontal lever A, Figs. 158, 159, is provided with a 
knife-edge b resting upon a strong vertical frame v; a jockey- 
weight w is movable along this lever and carries a vernier 
R by means of which the position of the weight can be read 
off upon a scale q. 

A second knife-edge c, carried by the lever, engages a link 
o connected to a cross-head. When operating for tension, 
one end of the specimen e is gripped in this cross-head, the 




s 



ZD 





r^ 


7 




irnr 




H 




K 


L 


m-. 


z^ 
, < 


G 




G 


r 1 






Fig. 159. 



Sf^ecimen 



other end being gripped in a cross-head l connected by rods 

G to an hydraulic ram F. 

If the resultant of the jockey- weight w and the lever A is 

W and acts at a distance y from the knife-edge B, we have by 

moments 

V .X = W 2/ 

X 



The scale Q is graduated so as to read off values of P direct, 
because W and x are of fixed value. 

Stops s are provided for the lever a, which normally rests 



366 



THE STRENGTH OF IVIATERIALS 



on the lower one ; as the pressure in the ram is increased the 
force exerted upon the specimen gradually increases until it 
reaches the value P, Avhereupon the lever rises and '"' floats " 
between the two stops. 

A lower cross-head j is suspended from the cross-head H by 
rods K and is used for compression and bending tests. The 
diagram on the right-hand side of Fig. 159 shows how the force 
is applied in the case of a compression test. In a bending 
test the arrangement is similar, but the test-beam is placed on 
supports on the cross-head J and a load point or points is or are 
connected to the cross-head L. 

The jockey- weight w is adjusted along the lever A by a 
screw which runs through the latter and is driven from a shaft 




Fig. 160.— Werder Testing Machine. 

O operated by a hand-wheel u or by power from a counter- 
shaft X. 

Werder Horizontal Single Lever IVIachine. — This 
machine is used to a great extent on the Continent, and is 
shown in diagrammatic form in Fig. 160. The lever a is of bell- 
crank type and the two knife-edges B c are close together so 
that the leverage is great and comparatively small weights w 
can be employed. The knife-edge B is carried by the hydraulic 
ram F, and the force p is transmitted to the specimen through 
a cranked lever D. It is quite clear from this diagram that as 
the s^^ecimen stretches the load would go off it if the ram did 
not follow, i.e. if the pressure Avere not maintained in the 
cylinder. When, as is common, this pressure is generated by 
a small hand-pump, the operator goes on pumping until the 
lever floats between the stops s. 

Compound Lever Machines. — Riehle Type. — Fig. 161 
shows a vertical type of compound lever testing machine, 




Fig. 161. — Riehle Testing Machine. 



[To jace paye 366. 



THE TESTING OF MATERIALS 



367 



made b}^ Riehle Bros., of Philadelphia, U.S.A., and used 
largely in America. 

The steelyard a is connected by a link with lever b, which is 
in turn connected with a lever c, which presses upwards upon 
the table or platen d. A cross-head E is operated by screws 
F, and according as the specimen is placed above or below this 
cross-head the test will be made in tension or compression. 
The machine shown is power driven by toothed gearing from 
an electric motor. 

These machines are controlled automatically by an electric 
contact device. At the outer end of the beam a are two 
contacts so arranged that when the beam reaches its highest 
position contact is made; this completes the circuit of an 



-S^ 





n ^^ 


G 




. ^rz 


D 


<^ 


1 1 


_ 


\ '^^ 


) 1 


9 






Sf:>€cimen. 






I _c 






1 1 


— 


^ — ' 


v^> 


//.'^y//'/^f^^ 


F 





Fig. 162. — Greenwood and Batley Compound Lever Testing Machine. 

electro-magnet which puts into gear with the driving 
mechanism the screw for moving the jockey -weight along 
the beam, but the movement of the jockey -weight can only 
follow up the extensions because contact is again broken as 
soon as the extension is more than is necessary to maintain 
the balance. Means are provided for varying the speed at 
which the weight is run out. An autographic recorder G is 
provided (see p. 379). 

Greenwood and Batley Horizontal Type. — This type 
of machine is made by Messrs. Greenwood and Batley, and 
was used by Professor Kennedy in the many researches which 
he carried out while at University College, London, this being 
one of the first testing machines installed in a college laboratory. 

The steelyard lever a, Fig. 162, has a knife-edge B and acts 
on a knife-edge c of a bell-crank lever d, which is pivoted upon 
a knife-edge e and is acted upon by a knife-edge F connected 



368 



THE STRENGTH OF JMATERIALS 



to a cross-head connected to the specimen. The other end 
of the specimen is carried by a cross-head operated by an 
hydraulic ram G. 

The usual leverage of the compound lever is 100 : 1 ; the 
jockey-weight w is generally moved along the steelyard, which 
carries a graduated scale, by means of a chain by a hand- 
wheel. 

WiCKSTEED-BucKTON HORIZONTAL Type. — This type is 
shown diagrammatically in Fig. 163 and by a photograph in 
Fig. 164. The steelyard lever a acts through a link c upon 
a bell-crank lever D, which connects by shafts shoT^n diagram- 
matically by G with the specimen. A massive carriage frame 




Fig. 163. — Wicksteed-Buckton Compound Lever Testing Machine. 

J is connected to the hydraulic ram r and carries a number 
of notches, into any of which can be fitted a cross- 
head K by which the other end of the specimen is carried. 
According to the position of the cross-head k, the specimen 
will be tested in tension or compression. This machine is very 
convenient for general testing on account of the ease with 
which it can be adjusted for different lengths of specimen and 
forms of test. 

Smaller Testing" Machines. — There are a large number 
of smaller testing machines in use, from which ver}' good results 
may be obtained in cases in which it is not essential for the 
specimens to be large ones. The student should remember 
that a great deal can be learnt with very simple apparatus. 
Fig. 165 shows a machine, designed by Professors Dixon and 
Hummel and manufactured by Messrs. W. and T. Avery, Ltd. ; 




&c 






o 



fa 



THE TESTING OF MATERIALS 369 

it has an automatic load-indicating device in the form of 
two dished plates connected by a patented flexible metallic 
diaphragm; the space between the plates is filled with a 
non-elastic fluid and the pressure is recorded upon a sensitive 
gauge which is graduated to give the load on the specimen. 
The gauge can be tested by means of a small plunger which can 
be loaded with weights supplied with the machine to produce 
pressures corresponding to the total capacity of the machine. 

The machine shown has a capacity of 10,000 lbs. and the 
force is applied by a capstan acting through worm and wheel 
gearing to a central screw. This gear can be thrown out for 
quick return and the screw operated direct by the handle shown. 

Calibration of Testing Machines. — To ensure accurate 
results in the use of testing machines they should be calibrated 
periodically ; the vertical type of machine possesses advantage 
in this respect because a heavy weight can be hung on direct. 

The first test is for zero error. This is effected by moving 
the jockey- weight carefully to the zero mark and seeing if 
the lever floats ; if it does not we can correct for this by an 
adjustment of the vernier on the jockey- weight by moving 
the latter until the lever floats and then moving the vernier 
until it reads zero. 

The next point that we may test is the value of the jockey- 
weight. This can be effected without removing it from the 
machine in the machines shown in Figs. 158, 163, by finding the 
floating position and then moving the jockey-weight a carefully 
measured distance I along the lever ; then at a distance z from 
the fulcrum suspend weights w until the lever floats again. 

iv z 
Then weight of jockey-weight = W = -y-. 

V 

To test for the accuracy of the knife-edge distance x we may 
proceed as follows : Hang a heavy weight Wi from the 
shackles of the machine and note the distance u that the 
jockey- weight has to move to balance it; 

then X = — 1 - 
w 

BB 



370 



THE STRENGTH OF :\L\TERIALS 



Another imjoortant test is for sensitiveness, by whicli is meant 
the amount by which the load may vary without causing the 
lever to come against its stops. This may be tested at zero 
in vertical machines by placing the jockey -weight at zero and 
hanging small weights on to the shackles until the lever ceases 
to " float " ; this should be repeated for larger loads and should 
also be tried by taking weights off as well as by putting them on. 

Grips and Forms of Test-Piece in Tension. — ^'^lien 
tests are made on flat bars, as is very common for rolled 
sections, wedge grips are generally' emj)loyed. Fig. 166 (a) 
shows one form of wedge grip. \Yedges A, pro^-ided with 
serrations to grip into the specimen, are driven into a tapered 




(«) 




Fig. 166. 



central passage through a block secured to the cross-head of 
the machine. 

In Fig. 166 (6) is sho^vn a grip suitable for a turned specimen 
provided with a collar. The collar bears against a washer c 
provided with a spherical end which bears against a tapered 
bush D, which engages in a similarly tapered central hole in 
the block B. This construction tends to keep the pull truty 
axial; a point of great importance. The ends of the speci- 
mens are very often screw-threaded, in which case they just 
screw into the blocks b. 

The British Engineering Standard Committee have specified 
the following rules, see Fig. 167, as to gauge length (cf. p. 55). 
(a) Flat Bars. — Gauge length = 8'" ; parallel for 9''. 

If the thickness is greater than J in., maximum width 
= IJ ins. 




Fig. 165. — Dixon and Hununel's Testing Machine. 

[To face page 370. 



THE TESTING OF MATERIALS 



371 



If the thickness is between f and J in., maximum width 
—■ 2 ins. 

If the thickness is less than f in., maximum width = 2 J ins. 
(6) Turned Sections. — Gauge length = 8 d; parallel for 9 d. 
(c) Turned Specimens from Forgings. — 

Area J in. ; gauge length = 2 ins. 
Area J in. ; gauge length = 3 ins. 
Area | in. ; gauge length = 3 J ins. 

Extensometers. — Extensometers are instruments for 
measuring the elastic strains of materials in tension or com- 
pression. In the types in most common use the strains are 



^ 



P 



— at 



8JL 



(b) 



(c) 

Fig. 167. 



^ 



~V 



magnified by an arrangement of levers and are measured by 
micrometer or by an indicator passing over a scale. We will 
describe a few of the most common types ; for other types 
a reference may be made to a paper by Mr. J. Morrow, in 
Proc. Inst. M. E. for 1904. 

An interesting report on the accuracy of various types 
of extensometers is given in the Report of the British Associa- 
tion for 1896. In these tests, different observers had bars of 
the same material sent for test. The results show very good 
agreement, some of the nearest results to the mean being 
obtained by instruments of very simple form. 

Goodman's Extensometer. — This extensometer is of very 
simple form and was designed by Professor Goodman, of 



372 



THE STRENGTH OE MATERIALS 



Leeds. It consists of two forked clips, a, b, Fig. 168, which 
carry pointed screws engaging in centre-punch marks in the 
specimen and are connected to rods c which join at their ends 
and carry a scale D on a projecting piece. 

Two light rods e, f form a fixed triangle, and the vertical 
rod E projects and has a small groove at its end which forms 
a bearing for a knife-edge carried by the pointer P. A second 
knife-edge on the latter rests upon a second vertical rod H 




Fig. 168. — Goodman's Extensometer. 

depending from the upper clip A. A small screw is pro- 
vided for bringing the pointer exactly to zero at the beginning 
of a test. The strain of the specimen causes the rod H to 
move slightly relatively to the rod e, this movement being 
magnified 100 times by the pointer lever. 

This and most other extensometers should be taken off the 
specimen as soon as the yield point is reached. 

Kennedy's Extensometer. — This extensometer was de- 
signed by Sir A. B. W Kennedy when professor at University 
College, London, and was one of the first lever extensometers. 
The instrument is for use in horizontal testing machines and 



THE TESTING OF MATERIALS 



373 




Fig. 169. — Kennedy's Extensometer. 

comprises two clips a^, Ag, Fig. 169, which carry triangular 
frames B^, Bg, which slide over and support each other. 



374 



THE STRENGTH OF MATERIALS 



The clips are as usual provided with pointed screws for 
engaging centre-punch marks in the specimen, lock-nuts being 
provided on the screws. As the specimen stretches, the 
frames slide relatively to each other, and a pointer-lever c 
which carries pins resting in depressions in each frame is thus 
caused to move over a scale d carried on an adjustable arm 




c 



Vj^m^ 



©B 



i^m^^j 



H 



c. J. r. Co. £-w. 



Fig. 170. — Ewing's Extensometer. 



E. To give an adjustment for zero, the depression in the front 
frame is formed in a plate r which can be adjusted by a fine- 
pitch screw. 

Ewing's Extensometer. — This instrument was designed 
by Sir J. A. Ewing when professor at Cambridge. 

The principle involved is illustrated diagrammatically in 
Fig. 170. There are two clips B and c each attached to the 
test-piece A by the points of two set-screws. The slip b has 



THE TESTING OF MATERIALS 375 

a projection b' ending in a round point P which engages with 
a conical hole in c ; when the bar extends this rounded point 
serves as a fulcrum for the clip c, and hence a point Q, equally- 
distant on the other side, moves, relatively to the clip B, 
through a distance equal to twice the extension. This dis- 
tance is measured by means of a microscope attached to the clip 
B . The microscope forms a prolongation of the clip B and the 
motion of the point Q is brought into the field of view by means 
of a hanging rod r. The rod R is free to slide on a guide in 
the chp B, and carries a mark on which the microscope is 
sighted. The displacement is read by means of a micrometer 
scale in the eye-piece of the microscope. The pieces B and b' 
are jointed to one another in such a way that the bar may twist 
a little, as it is sometimes liable to do during a test, without 
affecting the reading c. But the joint between B and b' forms 
a rigid connection so far as angular movement in the plane 
of the paper is concerned. This feature is essential to the 
action of the instrument : it is only then that P serves as a 
fixed fulcrum in the tilting of c by extension on the part of 
the specimen. 

Fig 171 is an illustration of the usual form of the complete 
instrument. The clips b and c in this standard pattern are 
set at 8 ins. apart. 

The object sighted is one side of a wire stretched horizontally 
across a hole in the rod b and illuminated by means of a small 
mirror behind. The distances cp and CQ are in this instance 
equal, with the effect that the movement of the sighted mark 
is double the extension of the test-piece. The length of the 
microscope is adjusted so as to give a constant magnification. 
This adjustment should be tested with the extensometer 
mounted on the specimen, and if necessary the length of 
the microscope tube can be altered by moving out or in the 
portion carrying the eye-piece. A complete revolution of the 
screw L, which has a pitch of -V of an inch, should cause a 
displacement of the mark through 50 divisions of the eye-piece 
scale, and when^this is the case the eye-piece is at the proper 



376 



THE STRENGTH OF MATERIALS 



distance from the objective. Readings are taken to tenths 
of a scale division, so that this displacement, which would also 
be given by ^i^ of an inch extension of the test-piece, 
corresponds to 500 units. Each unit then means ^,. ;.,,,. inch 
in the extension of the test-piece. 

A small extensometer based upon the same principle is 
used for measuring the compressive strain in short cylinders. 




Fig. 171. — Ewing's Extensometer. 

Dabwin's Extensometer. — This instrument has been 
designed by Mr. Horace Darwin, F.R.S., and is characterised 
by simplicity and solidity of construction, which make it 
suitable for heavy use. Another feature is that if the specimen 
should break unexpectedly when the extensometer is affixed 
little damage will result. 

The instrument is made in two separate pieces each of which 
is separately attached to the test-piece m, Fig. 172, by hard 
steel conical points P, P and p', p'. The steel rods carrying 
these points are mounted in] slides and after being driven 



__ 



THE TESTING OF MATERIALS 



377 



gently into the centre -punch mark in the test -piece are 
clamped in position by the milled heads r, r. Both parts of 
the instrument should be capable of rotating quite freely about 
the points, but there must be no backlash. 

The lower piece carries a micrometer screw fitted with a 
hardened steel point x and a divided head H. It also carries 
a vertical arm B at the top of which is a hardened steel knife- 




Fig. 172. — ^Darwin's Extensometer. 

edge. The upper and lower pieces work together about this 
knife-edge. A nickel-plated flexible steel tongue a forming a 
continuation of the upper piece is carried over the micrometer 
point X. This tongue acts as a lever magnifying the extension 
of the specimen, so that the movement of the steel tongue to or 
away from the steel point x is five times the actual extension 
of the specimen. 

To take a reading with the extensometer the thin steel 
tongue A is caused to vibrate and the divided head then turned 
till the point x just touches the hard steel knife-edge on the 



378 THE STRENGTH OF MATERIALS 

tongue as it vibrates to and fro. This has proved to be a most 
delicate method of setting the micrometer screw, and the 
noise produced and the fact that the vibrations are quickly 
damped out indicate to yoVo mm. the instant when the screw 
is touching the tongue. After the load is applied a second 
reading is taken in a similar manner and the difference in the 
readings gives directly the extension of the test-piece. 

If the test-piece is of small diameter the spring does not 
vibrate in so satisfactory a manner; the cause of this is the 
flexibility of the test-piece, the instrument itself vibrating as 
well as the spring. Still, very delicate readings can be taken 
by simply deflecting the spring with the finger and noting 
the contact as it passes the point. No damage can be done 
by advancing the micrometer screw too far forward ; all that 
happens is that the point passes the knife-edge on one side or 
the other. 

In the usual form, the gauge length is 100 mm. ; it may 
be pointed out that over the elastic portion of the test for 
which extensometers are used, the gauge length is not a matter 
of importance. 

Unwin's Extensometer. — This extensometer, designed by 
Professor Unwin, is shown in Fig. 173, and makes its measure- 
ment by a micrometer acting in conjunction with two spirit- 
levels. 

Two clips a, b, are secured to the test-bar by pointed set- 
screws, c, d, and carry sensitive spirit-levels g. The lower clip 
is first set level by means of an adjusting screw e ; the upper 
clip is then levelled by the micrometer screw /, on the graduated 
head of which readings are taken. When placed midway 
between the two edges of the specimen the extensometer gives 
the mean strain, but if placed to one side or the other by 
adjustment of the screws in the clips, the strain at any point 
in the width can be found in the case of eccentric loading. 

Morrow's Mirror Extensometer. — A simple extenso- 
meter enabling great magnification of the strain to be obtained 
is that designed by Mr. J. Morrow. Clamping screws a, b, 



THE TESTING OF MATERIALS 



379 



Fig. 174, pass through rings c, d, to the latter of which a vertical 
strip is rigidly attached, a pointed rod e acting as a distance 
piece. Between the ring c and the strip r is placed a small 
diamond-shaped prism h which carries a mirror m, a light spring 
clip s maintaining the requisite pressure between the prism 
and the ring c and strip f. A second mirror isr is attached to 




Fig. 173. — ^Unwin's Extensometer. 



the strip f and any change in length of the specimen will 
cause the mirror m to rotate relatively to the mirror isr. By 
observing the images of a scale in both mirrors, we obtain the 
strain by the difference of the two readings. 

Autographic Recorders. — Many testing machines are 
provided with mechanism for drawing the stress-strain (or 
more accurately the load extension) diagram automatically as 
the test proceeds. One of the earliest mechanisms of this 
kind was one used by Professor Kennedy upon a horizontal 



380 



THE STRENGTH OF MATERIALS 



compound lever machine. The movements of a pointer upon a 
piece of smoked glass were obtained in one direction by the 
actual extensions of the bar, and the movement represent- 
ing the stress was obtained by multiplying up the strain in a 




^ 



P; 



EL 



^^-£^^ 



f 



£^- 



E 



^ 






s 





tl 



Fig. 174. — Morrow's Extensometer. 



J^ 



longer bar coaxial with the test-bar, this longer bar being 
always stressed within the elastic limit and the load therefore 
being proportional to the extension. 

WiCKSTEED-BucKTON RECORDER. — This autographic re- 
corder is fitted to the Buckton machines and is shown on the 




Yia. 175. — Wicksteed-Buckton Autographic Recorder. 

[To face page 381. 



THE TESTING OF MATERIALS 381 

extreme right-hand side of Fig. 158, and also to larger scale 
in Fig. 175. The record sheet is placed on a drum, around 
which passes a string which goes through a tube t and passes 
between pulleys upon the cross-heads between which the 
specimen is gripped. As the specimen becomes strained the 
distance between these cross-heads varies and this motion is 
communicated to the drum so that the rotation of the drum is 
proportional to the strain. The stress is measured by first 
putting the jockey-weight w near its extreme position and 
preventing the lever a from coming down upon its stop by 
means of a spring s connected up to the pencil carriage. As 
the specimen becomes stressed, spring s is proportionally re- 
lieved from load and thus shortens in length by an amount pro- 
portional to the load appHed to the specimen. The upward 
movement of the pencil is therefore proportional to the load, 
and the combined movement of drum and pencil traces out a 
load-extension curve which is generally — ^though not quite 
accurately — called the stress-strain diagram. 

Torsion Testing Machines. — Single lever testing machines 
are often provided with an attachment for enabling testing 
by torsion to be carried out. Torsion tests to failure can be 
made upon comparatively small machines. 

Fig. 176 shows diagrammatically the form of testing machine 
used by Professor Kennedy; most other machines are based 
upon the same principle. A graduated lever a is counter- 
weighted to balance about a knife-edge B coaxial with the 
specimen x, which usually has the form shown in Fig. 167 (6) 
with the exception that the ends are not screw-threaded. 
A jockey- weight \v runs along the lever and the specimen is 
clamped in a chuck which is connected to the lever at the 
pomt c. The other end of the specimen is secured by a 
chuck carried by a worm-wheel d which is operated through 
a Avorm from a handle e to apply the necessary torque. The 
jockey-weight is placed so as to exert a given torque, and 
the handle e is turned until the lever " floats " between the 
stops s. 



382 



THE STRENGTH OF MATERIALS 



Professor Thurston's Torsion Machine. — In this 
machine, Figs. 177, 178, the specimen is a short one with 
square ends, one of which is carried by a jaw rotated by worm 
gear, the other being carried by a jaw connected to a weighted 
pendulum, the angular movement of which determines the 
torque applied. 




X~ 



^ 



Fig. 17G. — Torsion Testing Machine. 

An autographic diagram is obtained by securing a pencil 
to the pendulum in such a manner that the pencil moves 
parallel to the axis of the specimen as the pendulum swings 
outwards. A cylinder carrying a paper strip is secured to 
the jaw carried by the worm-wheel. The paper thus rotates 
by an amount equal to the angle of torsion, and the pencil 
moves at right angles by an amount which is a measure of the 
torque applied. 



THE TESTING OF MATERIALS 



383 



A templet of the form shown in Fig. 178 is employed to 
obtain a standard size of specimen. 

Avery's Reverse Torsion IMachine. — Fig. 179 shows two 
views of a torsion machine, patented by Messrs. Avery, by 
means of which a torque can be applied in either direction. 




Fig. 177. — Thurston's Torsion Testing Machine. 

The specimen x is gripped between special three- jaw chucks 
g, g', and the torque is applied from a hand-wheel i through a 
worm-wheel h mounted upon an adjustable standard a*. The 
torque is thus communicated to a lever / and thence through a 
supplementary lever k and rod I to the steelyard b, upon which 
the usual jockey d is mounted. 



384 



THE STRENGTH OF MATERIALS 



The main torsion lever / is fulcrummed on ball-bearings /^. 
Passing through the centre of this bearing is a spindle e of cruci- 
form section to which the chuck g is attached. This spindle 
is connected with the lever / by means of rollers e^ whereby 
it has limited longitudinal movement through the lever, but 
cannot revolve therein. This longitudinal distance is to allow 




Iflb 



Fig. 178. 



of adjustment due to the shortening of specimens undergoing 
tests, the collar e^ on the spindle e preventing the withdrawal 
or extended movement of the spindle. The main lever / is 
provided with knife-edges p and /^ through Avhich connection 
is made to the supplementary lever k ; this lever k is w ithin 
the main lever / and has a ball-bearing fulcrum k^ on a bracket 
a^. It is suspended by means of the link m from the knife- 
edge /2 of the main lever, and at its opposite end it is connected 
with the knife-edge of the main lever P by a link m^. The links 
m and m^ connect with the lever k through Imife-edges k^ and 
P. Between the knife-edge P and the ball-bearing k^ is 
another knife-edge n which forms the connection from the 
levers / and k to the steelyard b by means of the rod Z. The 



THE TESTING OF MATERIALS 



385 



lever / is counterbalanced by the adjustable weight /*, and the 
lever ^ by a similarly arranged weight k^. 

Assuming the torque to be applied as indicated by the arrow 
y, it lowers the end of the lever / which is in contact with the 





^>ci 






^1 


--------- 





Fig. 179.— Avery's Reverse Torsion Machine. 

lever k ; the point of greatest depression in contact with the 

lever k will be the knife-edge /^ which through the link m^ and 

knife-edge k^ lowers this end of the lever k, about its fulcrum k^, 

so that this movement of the lever k lowers the knife-edge n, 

thereby exerting a downward pull on the connecting rod I and 

raising the free end of the steelyard b. 
cc 



386 THE STRENGTH OF MATERIALS 

If the torque is a^oplied in the direction indicated by the 
arrow z the end of the lever / which is in contact with the lever 
Ic is raised and the knife-edge /^ also raised ; this upward move- 
ment of ths knife-edge /^ raises the link m dependent there- 
from and also the knife-edge k^ of the supplementary lever A:, 
causing the lever Ic to move about its fulcrum 1^ as before. 
The knife-edge n is again depressed by this movement of the 
lever and exerts as before a downward pull on the connecting 
rod. By this arrangement, in whichever direction the tor- 
sional stress is applied to the main lever /, the resultant 
direction of force on the connecting rod I is the same. 

Professor Lilly's Reverse Torsion LIachine. — This 
reverse torsion machine, patented by Professor Lilly of 
Dublin, is a very simple machine for obtaining autographic 
diagrams in torsion and is particularly of value when working 
within the elastic limit. 

A circular table a. Fig. 180, is fixed to any convenient bench 
or stool, and has through its centre a hollow steel cylinder B. 
In the central part of the cylinder is placed the specimen c to 
be tested, one end being secured to it by the key at d and the 
other end passing through the adjustable bearing E ; it is 
secured to the lever g H i by the key at F. The lever consists 
of a solid shank h, which is rigidly connected to the spring i ; 
the weight G with its connecting arm forms part of the solid 
shank, and is for the purpose of balancing the lever. Fixed 
to the spring i at J is a light arm k, at the end of which is an 
adjustable spring Q carrying the recording pencil l. This 
pencil is adjusted to slide along the straight edge N which 
forms part of the frame s. A circular drum m revolves on its 
outer edge o on the table A, and is connected by adjustable 
pivot bearings to the frame s which is connected by adjustable 
pivot bearings R to the arms of the solid shank h. 

The manner of carrying out a torsion test with the machine 
is as follows : The specimen c to be tested is placed in 
position in the cylinder B, and secured to it by driving up 'the 
key at d ; the lever G h i is now placed in position on the top 



THE TESTING OF MATERIALS 



387 



end of the specimen c and secured to it by driving up the key 
at F. A sheet of squared paper is fixed on the drum m, which 







o 
H 



> 
P5 



o 

M 

(D 
O 



o 

00 






is then put in position on the table a, and the pivot bearings 
adjusted; the pencil l is now placed in the central position 



388 THE STRENGTH OF MATERIALS 

on the drum and in contact with the straight edge n by adjust- 
ing the arm k and the spring q. The torsion test on the 
specimen is carried out by applying a pull or push to the handle 
p; the pencil L then automatically graphs the stress-strain 
diagrams on the squared paper. The movement of the pencil 
L along the straight edge n is proportional to the push 
or pull on the handle p and gives to scale the magnitude 
of the torsional or twisting moment ; this may be shown as 
follows — 

Regarding the spring i as a cantilever with a load at the free 

(l ij 
end, the value of y and -—- at the point J is proportional to the 

force applied at p. Calling the length of the arm z, the deflec- 

d u 
tion of the pencil end is y ~ z ^^ which is proportional to the 

applied force. The roll of the drum m under the pencil is 
proportional to the angle of torsion of the specimen. The 
pencil graphs the combination of these two movements at 
right angles to one another, and the resulting stress-strain 
diagrams are thus obtained to rectangular co-ordinates. 

To calibrate the machine a known. pull is applied to the 
handle p by means of a spring or otherwise, and the dis- 
tance traversed by the pencil along the straight edge gives 
to scale on the squared paper the magnitude of the apphed 
twisting moment. The scale of the angle of torsion is obtained 
by observing the number of turns of the drum during one 
complete revolution on the circular table. 

For examples of diagrams taken with this machme the 
reader is referred to a paper in Froc. Inst. C. E. Ireland, 
Vol. 41. 

Professor Coker's Combined Bending and Torsion 
Machine.* — This machine has been devised by Professor Coker 
for experiments upon combined bending and torsion. 

The various parts are supported in a built-up frame con- 
sisting of two planished steel shafts, a, Fig. 181, secured in 

* Phil. Mag., April 1909. 



THE TESTING OF MATERIALS 



389 



cast-iron cross frames b, mounted on four standards, one 
of which latter is adjustable in height to secure steadiness 
on an uneven floor. U23on the steel shafts are two castings c, D, 
each of which has a cylindrical bearing E encircling one of the 
shafts and resting with a flat face F in line contact with the 
other shaft, and secured in j)osition by a cross-bar G threaded 
on studs. This connection is perfect!}^ rigid, since it removes 
all degrees of freedom, and it is readily released by simply 
turning back one of the cross-bar nuts, leaving the casting 
free to slide into a new position. It also has the advantage 





Fig. 181. — Coker's Combined Bending- and Torsion Machine. 



that no accurate fitting is required for the supporting frame. 
The casting c carrying the worm-wheel gear w has trunnion 
bearings h at right angles to and intersecting the axis of the 
specimen. The bearings are fitted with friction rollers, and 
when the machine is used simply for torsion the worm-wheel 
is kept in a vertical position by an arm i keyed to the bearing 
H and locked in position by a thumb -screw. A weight J 
attached by an arm to the second bearing balances the pivoted 
casting in all positions. 

The weigh-levers are supported from a vertical standard 
K of the frame d by a wire l, terminating in a thin plate M 



390 THE STRENGTH OF MATERIALS 

with a keyhole slot encircling the spindle N. Formerly a 
roller bearing was used for this spindle, but this is an un- 
necessary rejBnement, as the friction is extremely small and 
can be easily taken into account. The casting supported 
in this way has three levers, p, q, and r, the first two of 
which are for the application of twisting moments s, and the 
third R, in the line of the specimen, is for applying a bending 
moment. 

All the loading levers are provided with knife-edges of 
circular form, made by turning an ordinary Whitworth nut 
down to form a disk with a V-shaped edge. These disks 
carry rings t with wdde-angled V-shaped recesses on the inner 
sides, and light rods v screwed into these rings carry the 
weights. This arrangement of knife-edge is very easy to 
adjust accurately, and when bending and twisting stresses 
are applied simultaneously the rolling line contact adjusts 
itself to the bending and twisting of the specimen. The 
bending of the specimen causes a change in the effective 
arm of the bending levers, which is generally negligible, but 
a correction may be necessary wdth a very long specimen. 
For if a is the length of the lever-arm, and h is the radius of 
the circular knife-edge, an angular deviation of amount 
will cause a change of a — {a cos ^ + 6 sin 0) in the lever- 
arm, and this is zero when ^ = and also when a = a 
cos ^ -f 6 sin ^. 

In the machine described a is 10 inches and h is 0*5 inch, 
and the angles ^ = and 6 = 515° both correspond to an 
effective length of 10 inches. The maximum correction be- 
tween these values is easily shown to be at an angle given 
by the equation b cos = a sin 6, in the present case 29° 
approximately, for which value the correction is 0*12 per 
cent. In the majority of tests the angular change at the 
ends rarely exceeds 5°, and the correction is therefore so very 
small as to be practical!}^ negligible. 

The worm-wheel w and the casting v for the weigh-levers 
are bored out to receive the ends of the specimen, and are 



THE TESTING OF MATERIALS 



391 



provided with fixed keys which slide in corresponding key- 
ways cut in the specimen. When tubes are subjected to stress 
they are provided with solid ends secured by transverse pins, 
thereby avoiding brazed joints, since these latter are trouble- 
some owing to the state of the metal being altered by the 
brazing. The end of the specimen projecting through the 
worm-wheel is fitted with a lever x for applying bending 
moment, and both levers for bending may be loaded in- 
dependently or by a cross-bar suspended from stirrups as 
shown. 



CaPiefomelgr 




Fig. 182. — Simple Torsion Meter. 



Torsion Meters. — The elastic angular strain in torsion 
requires less magnification than the elastic longitudinal strain 
in tension, and so comparatively simple apparatus can be 
used. 

Fig. 182 shows a simple apparatus made by Mr. A. Macklow- 
Smith. Two arms a, b, connected together by an extensible 
sleeve, are secured by pointed screws to the specimen. The arm 
A carries a cathetometer or telescope and the arm B carries 
an ivory scale upon which the angle of torsion is read. 

Professor Coker's Torsion Meter.* — This apparatus, 

* Phil. Mag., April 1909. 



392 



THE STRENGTH OF MATERIALS 



which can be used also for measuring the strain in combined 
bending and torsion, consists of a graduated circle A, Fig. 183, 
mounted on the specimen b by three screws c in the chuck- 
plate D. A sleeve e provided with three screws grips the 
specimen at a fixed distance away from the first set. The 
spacing of these two main pieces on the specimen is effected 
by a clamp, not sho^Mi in the figure, which grips the double 
ones P, G, and maintains them at the correct distance apart 
until the set -screws are adjusted. 




Fig. 183.— Coker's Torsion Meter. 



The clamp is afterwards removed, leaving the plane of the 
graduated circle perpendicular to the axis of the specimen and 
the sleeve correctly set and ready to receive the reading 
microscope H. 

The vernier plate carries a sliding tube i, on which a wire j 
is mounted, and the movement of the latter due to bending 
or twist is measured by a scale in the eye-piece K, the divisions 
of which are calibrated by reference to the graduated circle. 
It is found convenient to have the microscope-tube pivoted 
about an axis perpendicular to its central line at L, so that 
any slight difference due to imperfect centring can be adjusted 



THE TESTING OF MATERIALS 393 

by the screw m to make the calibration value agree for a series 
of specimens. 

Torsion Dynamometers (or torsiometers as they are 
sometimes called) are instruments for indicating the horse- 
power being transmitted by a shaft rotating at a known 
speed by measuring the angle of torsion over a given length. 
They are often provided with autographic record devices. 
The horse-power is derived by a combination of formula (3), 
p. 312, and formula (11), p. 324. 

Repetition Stress Machines. — We described on p. 85 one 
of the forms of machine used for rotating beams by Wohler 
in his experiments in repetition of stress. Similar machines 
are in use in many engineering laboratories ; in most cases 
the spring is replaced by a weight which has a spherical socket 
resting on a spherical bearing fixed to the end of the specimen. 
By this arrangement the weight remains free from oscillation 
as the specimen sags under load. 

Professor J. H. Smith's Machine. — In this machine, which 
is shown in Fig. 184, the variation of stress is caused by the 
variation of the longitudinal component of the centrifugal 
force of the rotating weights e. 

The specimen which is to be tested connects the two pieces 
c and B. The upper piece is of circular cross section, and has 
a locking arrangement consisting of a cap and set screws. 
The lower piece b is of circular section in the upper bearing l, 
and of square section in the lower bearing m. 

The two pieces c and b are supported by the frame- 
work F of the machine; and the specimen is inserted with- 
out straining it by first locking it to the piece b, then by 
locking c to the specimen and afterwards locking c to the 
frame. 

The piece b has a bearing n at right angles to its length 
in which a spindle revolves; there are two plates k, k, and 
weights E, e, rigidly attached to this spindle. 

The rotating spindle is driven by means of a pair of pieces 
in contact, one, a crank pin diametrically opposite to e, on 



394 



THE STRENGTH OF MATERIALS 



one of the rotating plates K, and one a radial slot on a plate 
connected with either another unit or to a shaft rotating in 
fixed bearings. By this arrangement the driving force is not 
transmitted to the specimen. 

The component of the centrifugal force exerted by the 





Fig. 18-i. — Smith's Stress Repetition Machine. 



rotating weights e, e, produces an alternating stress in tlie 
specimen. 

The spring h, and tightening device j, enable the operator 
to put any desired amount of tension or compression in 
the specimen. This spring may be replaced by weights and 
levers or by an hydraulic cylinder. The lead rings or springs 
at G, G, act as buffers and receive the blow when the specimen 
breaks. 



THE TESTING OF MATERIALS 395 

The complete machine consists of one or more units together 
with the necessary balancing rotating masses which may be 
either parts of other units or parts connected to a revolving 
shaft mounted on the framework. 

See also Professor Arnold's machine, p. 399. 



CHAPTER XIV 

THE TESTIIS'G OF MATERIALS {cOTltd.) 

Impact, Ductility, and Hardness Testing. — As we 
have indicated on p. 54, the elongation under a tensile test is 
commonly taken as a measure of the ductilit}-, but experience 
shows that this is not sufficient in all cases, and in recent years 
a number of simple machines have been devised for carrying 
out tests upon small specimens. 

Cold and hot bending tests are commonly specified by the 
various authorities and purchasers of steel and iron, giving the 
angle through which the specimen must bend without cracking. 
In the specification for structural steel issued by the British 
Standards Committee, for instance, there is a clause that test- 
pieces must without fracture withstand being doubled over 
until the sides are parallel and the internal radius is not greater 
than IJ times the thickness of the test -piece, the latter being 
not less than IJ inches wide. 

Tests of this kind have the advantage that they can be 
made in the workshop without special apparatus, but the 
disadvantage that the results are rather negative. 

Captain Sankeys Hand Bendi>'g Machixe. — In this 
machine, patented by Captain Sankey, a j)iece of metal is 
bent backwards and forwards through a fixed angle until it 
is broken ; the bending moment being measured by the 
deflection of a spring and recorded upon a paper drum. 

The standard angle is 9H°, i.e. 16 radians, so that the work 

done to make a complete bend is obtained b}' multiplying the 

bending moment b}' 16. 

396 



THE TESTING OF MATERIALS 



397 



At one corner of the base of the machine there is a grip A 
for securing one end of a flat steel spring b . The other end of 
the spring is fitted with a special grip c for holding one end 
of the test-piece d. The other end of the test-piece is fixed 
into a handle E, about 3 feet long, by means of which it is 




Fig. 185. — Sankey's Hand Bending Testing Machine. 

bent backwards and forwards through the standard angle. 
A graduated arc F is provided to show this standard angle. 
Alongside of the spring, and fixed to the bed-plate, there is a 
horizontal drum g, to carry the recording paper, and the 
pencil H has a horizontal motion actuated by the motion of 
the grip c and conveyed by the steel Avires l and m and the 
multiplying pulley n, the wires being kept taut by a weight. 
The zero line is in the middle of the paper, and the pencil H 



398 THE STRENGTH OF MATERIALS 

moves in one direction when the bending is from right to left 
and in the opposite direction when it is from left to right. The 
drum has a ratchet wheel K, with a detent (not shown) 
worked by the motion of the pencil carrier. The result of 
the combined motion of the pencil and of the drum is to 
produce an autographic diagram such as shown. Obviously, 
the greater the stiffness of the test-piece the more the flat 
spring B will have to be bent before its resistance is equal 
to the resistance to bending of the test-piece. Hence the 
motion of the pencil is proportional to the bending moment 
required to bend the test-piece. 

In operation, the test -piece is properly secured in the handle 
E by means of the set screw ; it is then inserted into the grip 
c, and the free length (1| inches) is adjusted by means of a 
gauge provided for the purpose, after which the grip c is 
tightened. The handle is slowly pulled towards the left until 
the specimen is felt to be " yielding " — this action can be 
distinctly felt, and this bend is known as the " yield bend." 
Without altering the pressure on the handle, the record 
cyHnder is now rotated two teeth by working the detent by 
hand, and the first bend is completed by making the mark on 
the handle coincide with the pointer indicating the " standard " 
angle. The bending is then reversed, and the test-piece is 
bent until the mark on the handle coincides with the second 
pointer. The bending is again reversed, and so on until the 
specimen breaks. The point at which the test-piece breaks 
should be noted in decimals of one bend, which are marked on 
the graduated arc. 

The machine is cahbrated by fixing the handle end of the 
lever in the jaws and applying a known force by means of 
a spring-balance and comparing the record made on the strip 
with the actual moment appHed. If there is any discrepancy 
between the two results, the spring is adjusted until such 
discrepancy disappears. The number of bends which a give^ 
material can endure before fracture is a measure of the 
ductihty, and experiment shows that this is approximately 



THE TESTING OF MATERIALS 



399 



proportional to the percentage elongation multiplied by the 
percentage reduction in area in a standard tensile test. 

The following empirical results have been found from 
experiment to be approximately true — 

Yield-point stress in tons per sq. in. 

_ First bending moment in lb. ft. 

~ r55 

Ultimate tensile stress in tons -per sq. in. 
_ Longest line in lb. ft. 

^ r55 




Fig. 186. — Arnold's Testing Machine. 



The energy required to cause rupture is equal to 1'6 multi- 
plied by the number of bends, multiplied by the mean range 
of bending moment in lb. ft. 

Professor Arnold's Reverse Bending Machine. — In 
this machine, which may also be regarded as a repetition of 
stress machine, a bar a, Fig. 186, f in. in diameter, is firmly 
held in a clamp b and passes through a slot in a slide c which 
is reciprocated by a shaft d running at a standard speed of 
650 revolutions per minute. The distance between line of 
contact of the slot with the specimen and the point where the 



400 



THE STRENGTH OF rilATERIALS 



latter enters the clamp is 3 inches and the slot is adjusted to 
cause a deflection of | in. in the specimen on each side. The 
number of bends which the sj^ecimen endures before fracture 
is taken as a measure of the capacity of the material to resist 
failure by shock. 

Repeated Impact Testing Machine. — The machine 
shown in Fig. 187 is made by the Cambridge Scientific Instru- 
ment Co., Ltd., and is a modification of a machine described 
by Messrs. Seaton and Jude * and used by Dr. Stanton j of 
the National Physical Laboratory. 



.-1 

--1 




,z:L 



■«i -r-- 






J s 




-^ 



Mr 



M 



Fig. 187, — Repeated Impact Testing Machine. 

The machine is fitted with a cone-pulley a, so that it can be 
driven by a belt from a hne shaft or small electric motor. One 
end of the spindle driven by this cone-pulley carries a crank b 
which is connected to the lifting rod c. This lifting rod is 
supported on a roller d, at some point in its length, so that the 
circular motion imparted to the rod at the crank end causes 
it to rock and shde on the roller. Thus an oval path, shown 
in dotted lines, is traced by the free end of the lifting rod. At 
this end the rod is bent at right angles so that on the upstroke 
it engages with and hfts up the hammer head e. This hammer 
head is fixed to the rod f, which is hinged at the end g. 
Having reached the top of its path, the lifting rod c moves 
* Froc. I. Mech. E., 1904. t Proc. I. Mech. E., 1908. 




Fig. 188. — Impact Testin.o- Machine. 



[To face page 401. 



THE TESTING OF MATERIALS 401 

forward and disengages the hammer, which then falls freely 
on to the specimen h under test. 

This cycle is repeated from 70 to 100 times a minute. The 
height through which the hammer falls can be varied by moving 
ing the roller d along a scale m which is calibrated to read 
directly the vertical height through which the hammer falls. 
Adjustment can be made by this means up to a maximum of 
3 J inches (90 mm.). 

The specimen h is usually about J^' (12 mm.) in diameter, 
with a groove turned in it at its centre to ensure its fracture 
at this point in its length. It is supported on knife-edges 4 J'' 
(114 mm.) apart, the hammer striking it midway betw^een these 
knife-edges. The knife-edges are cut slightly hollow, and a 
finger spring holds one end of the specimen in place. The 
other end is held in a chuck which is hinged in such a manner 
that it does not take any portion of the hammer blow, all of 
which comes on the knife-edges. 

The specimen remains stationary whilst the blow is struck, 
but between the blows it is turned through an angle of 
180°. 

A revolution counter to register the number of blows struck 
is fixed to the bed plate of the instrument. When fracture 
occurs, the specimen falls away, and the hammer head 
continues to fall, first tripping an electric switch, and finally 
coming to rest on a steel stop-pin h. 

Izod's Impact Testing Machine. — This machine, which 
is made by Messrs. Avery, tests the impact-resisting qualities 
of a material by measuring the energy absorbed from a pendu- 
lum which breaks a projecting notched specimen as it swings 
past it. 

The specimen is 2 inches long, -f^ inch thick and f inch broad, 
and has a notch cut in it by means of a templet ; it is held in 
the \ace shown at the base of the machine. Fig. 188, and the 
pendulum is then released from a fixed height by means of a 
trigger. The energy required to fracture the specimen takes 
some of the swing out of the pendulum, and the height to which 

DD 



402 THE STRENGTH OP MATERIALS 

the latter swings on the other side is indicated by the pointer 
passing over the scale at the top of the specimen. The scale 
is graduated to give the energy directly in foot pounds. A 
brittle material will not absorb much of the energy, whereas 
a tough material will absorb a good deal. 

Inasmuch as the base is not absolutely rigid, the results of 
tests in this machine are relative rather than absolute, but 
it gives very useful results in practice and has the advantage 
that the tests can be made in a very short time. 

The same machine can be adapted for testing hardness by 
impact. A strong cast-iron anvil is provided in which a 
specimen 1 inch in diameter and 1 inch long is placed. The 
pendulum strikes a loose plunger which carries a hardened 
steel ball; this ball is placed in contact with the specimen 
prior to the impact, which causes an indentation in the 
specimen. The diameter of this indentation is taken as a 
measure of the hardness as in the Brinell machine next to 
be described. 

Brinell's Hardness Testing Machine. — In this machine 
a hardened steel ball is pressed with a predetermined force 
against the plate whose hardness is required. The diameter 
of the resulting curved depression is then found and from this 
the " hardness number " is obtained in the manner described 
below. 

Fig. 189 shows the Brinell machine made by Messrs. J. W. 
Jackman «fe Co., Ltd. The specimen is placed upon the top 
• of the stand, which is then adjusted by the hand-wheel to 
bring the specimen into contact with the hardened steel ball 
(10 mm. diameter) which projects from the conical end of the 
plunger of the machine. The upper portion of the machine 
comprises a small fluid-operated testing-machine, oil being the 
working fluid. By means of a small projecting pump-handle 
the pressure of the fluid is increased until the cross-piece 
" floats," the pressure being indicated on the dial. Weights 
are provided with the machine, which make the floating occur 
at a force of 500 to 3000 kg. (increasing 500 kg. at a time). 




'~1 



Fig. 189. — Brinell Hardness Testing Machine. 

[To face page 402. 



THE TESTING OF MATERIALS 



403 



The pressure depends only on the weight applied and not 
upon the accuracy of the gauge. 

If P is the load, D the diameter of the ball and d that of the 
impression, the quantity 

H = 



D _ V D2 - d^ 



D 



is called the Brinell Hardness Number. The following table 
gives values. 

BRINELL'S HARDNESS NUMBERS (for load 3000 kg.) 
Diameter of Steel Ball =10 mm. 



Diameter 




Diameter 




Diameter 




Diameter 




of 


Hard- 


of 


Hard- 


of 


Hard- 


of 


Hard- 


Ball Im- 


ness 


Ball Im- 


j ness 


Ball Im- 


ness 


Ball Im- 


ness 


pression 


Number 


pression 


Number 


pression 


Number 


pression 


Number 


mm. 




mm. 




mm. 




mm. 




2-0 


946 


3-25 


351 


4-50 


179 


5-75 


105 


2-05 


898 


3-30 


340 


4-55 


174 


5-80 


103 


2-10 


857 


3-35 


332 


4-60 


170 


5-85 


101 


215 


817 


3-40 


321 


4-65 


166 


5-90 


99 


2-20 


782 


3-45 


311 


4-70 


163 


5-95 


97 


2-25 


744 


3-50 


302 


4-75 


159 


6-0 


95 


2-30 


713 


3-55 


293 


4-80 


156 


6-05 


94 


2-35 


683 


3-60 


286 


4-85 


153 


6-10 


92 


2-40 


652 


3-65 


277 


4-90 


149 


6-15 


90 


2-45 


627 


3-70 


269 


4-95 


146 


6-20 


89 


2-50 


600 


3-75 


262 


5-0 


143 


6-25 


87 


2-55 


578 


3-80 


255 


5-05 


140 


6-30 


86 


2-60 


555 


3-85 


248 


5-10 


137 


6-35 


84 


2-65 


532 


3-90 


241 


5-15 


134 


6-40 


82 


2-70 


512 


3-95 


235 


5-20 


131 


6-45 


81 


2-75 


495 


4-0 


228 


5-25 


128 


6-50 


80 


2-80 


477 


4-05 


223 


5-30 


126 


6-55 


79 


2-85 


460 


4-10 


217 


5-35 


124 


6-60 


77 


2-90 


444 


4-15 


212 


5-40 


121 


6-65 


76 


2-95 


430 


4-20 


207 


5-45 


118 


6-70 


74 


30 


418 


4-25 


202 


5-50 


116 


6-75 


73 


3-05 


402 


4-30 


196 


5-55 


114 


6-80 


71-5 


3- 10 


387 


4-35 


192 


5-60 


112 


6-85 


70 


315 


375 


4-40 


187 


5-65 , 


109 


6-90 


69 


3-20 


364 


4-45 


183 


5-70 1 


107 


6-95 


68 



For other test loads, the hardness numbers are proportional 
to those in the table. 



Within certain limits the Brinell Hardness Number of a 



404 



THE STRENGTH OF MATERIALS 



material gives a very fair indication of its tensile strength. 
Thus for steels with a hardness number less than 175, the 
ultimate tensile stress in tons per sq. in. is obtained approxi- 
mately by multiplying the hardness number by -23. 



TESTING CEMENT AND CONCRETE 

Tension Tests. — The form of briquette in accordance 
with the Specification of the British Engineering Standards 
Committee is shown in Fig. 190, the cross-section being 1 sq. in. 
at the weakest point. We have given on p. 78 the require- 




FiG. 190. — Cement Briquette. 

ments as to tensile strength in accordance with this specifica- 
tion. As the strength obtained under test is found to depend 
upon the rate of loading, being higher for quick loading, the 
above specification stipulates that the loading shall be at a 
rate of 500 pounds per minute. 

A simple form of lever machine, made by W. H. Bailey 
& Co., is illustrated in Fig. 191. The specimen is gripped in 
the shackles and the load is applied by allowing shot to fall 
into the bucket, the leverage being such that the tension 
applied is fifty times the weight of the shot. The shot-hopper 
is provided with a valve, the operating arm of which passes 
over the lever, so that when the specimen breaks the supply 
of shot is automatically cut off. The shot is then weighed, a 



THE TESTING OF MATERIALS 



405 



spring-balance being often used which gives readings equal to 
fifty times the weight of the shot, thus giving the breaking 
stress direct. 

A rather more accurate form of machine, made by the same 
firm, is shown in Fig. 192. In this machine water is allowed 
to run slowly into a long graduated can placed at the end of 
the lever. The supply of water is cut off when fracture occurs 




Fig. 191. — Cement Testing Machine (Tension). 

and a gauge glass placed outside the can is provided with a 
scale graduated to enable the breaking stress to be read off 
direct. 

In another common form of testing machine a jockey-weight 
is moved automatically along a lever arm by means of a weight 
controlled by an adjustable dashpot which enables the rate 
of loading to be varied. On the fracture of the specimen the 
weight becomes stationary and the breaking stress is read off 
on a scale attached to the lever. 

Compression Tests. — Compression tests are not usually 



406 



THE STRENGTH OF MATERIALS 



specified for pure cement, although they are becoming more 
common. For concrete in reinforced concrete works, however, 
compression tests are nearly always required. 

A common specification is that cubes, the area of each side 
being 50 sq. cm., of 3 parts sand to 1 part cement by weight 




Fig. 192. — Cement Testing Machine (Tension). 



shall develop at 28 days at least 10 times the standard tensile 
strength (^. e. 2000 lbs. per sq. in.). 

The following test results for the concrete cubes (with area 



THE TESTING OF MATERIALS 



407 



of each side 50 sq. cm.) are recommended by the Concrete 
Institute. 



Proportion by Volume. 


Crushing Strength in lbs. 
per sq. in. 


Cement 


Sand 


Coarse 
Material 


28 days after 
mixing 


120 days after 
mixing 


1 

1-2 
1-5 
2 


2 

2 
2 
2 


4 
4 
4 
4 


1600 
1800 
2000 
2200 


2400 

2600 
2800 
3000 




Fig. 193. — Compression Tension Press for Cement, etc. 

A common form of special machine for crushing tests of 
cement, concrete, etc., is shown in Fig. 193. The cube is first 
fixed by means of the upper hand- wheel and the side -press 



408 THE STRENGTH OF MATERIALS 

screw is then operated to compress the operating fluid (oil 
or glycerine). The crushing pressure is recorded by the 
pressure gauge which is constructed so as to maintain its 
reading after fracture has occurred. 

Specific Gravity, Fineness, Soundness, and Setting 
Tests for Portland Cement. 

Specific Gravity. — According to the British Standard 
Specification, the specific gravity of Portland cement shall not 
be less than 3* 15 when fresh or 3* 10 after 28 days from grinding. 
• There is considerable doubt as to the value of this test; 
some very useful information on this and other matters in 
cement testing will be found in a paper on " Common Fallacies 
in Cement Testing," read by Mr. W. L. Gadd, F.I.C., before 
the Concrete Institute, December 11, 1913. A chemical 
analysis appears to be a much more reliable test. 

Fineness. — The fineness test is applied by means of standard 
sieves. In the British Standard Specification not more than 
18 % residue is allowed for a sieve of square mesh with 180 
wires, each '002 in. in diameter, per inch, and not more than 
3 % for a 76 mesh with wire '0044 in. in diameter. 

Le Chatelier Soundness Test. — This test is conducted 
with a piece of split brass tube a, Fig. 194, 30 mm. internal 
diameter and 30 mm. long, the thickness of metal being J mm. 
Pointers b b are attached to the tube, the length from their 
points to the centre of the tube being 165 mm. 

This tube is used as a mould, and is filled with wet cement, 
one end being placed previously on a piece of glass ; the other 
end is then covered with a weighted piece of glass and the 
whole is placed in water at a temperature from 58° to 60° F. 
and left for 24 hours. The distance between the pointers is 
then measured and the mould is then placed in water which 
is heated to boihng point and maintained in that condition 
for 6 hours. The British Standard Specification stipulates 
that after boiling the increase in the distance apart of the 
pointers shall not be greater than 6 mm. 

Setting Tests. — Setting tests are often made by finding 



THE TESTING OF MATERIALS 



409 



the time before a standard weighted needle fails to make 
an impression in the cement. The Standard Specification 
requires that before any sample is submitted to setting test 
it shall be spread out for a depth of 3 inches for 24 hours in a 
temperature from 58 to 64° F., and that the setting time shall 
not be less than 2 hours nor more than 7 hours. Mr. Gadd in 



B' 



■B 




Fig. 194.— Chatelier Cement Test. 

the paper referred to above has shown that this spreading 
for the purpose of aeration leads to very variable results, 
depending largely on the locality and humidity. 

The Thermal Method of Testing Materials. — It was 
apparently first noticed by Magnus that changes of stress 
are accompanied by a change of temperature ; when a body 
is stretched its temperature lowers very slightly, and when 
it is compressed or squeezed its temperature rises. By 



410 THE STRENGTH OF MATERIALS 

means of a thermopile and a very delicate galvanometer, 
therefore, the changes of temperature at various points of a 
structure, and therefore the stresses, can be determined. Care 
should be taken to distinguish the phenomenon under con- 
sideration from the well-known heating that occurs at the 
yield point in tension experiments ; this is very much greater 
in magnitude and is reverse in sign, the elastic tension being 
accompanied by a fall in temperature. 

Lord Kelvin has deduced the following formula to deal 
with the problem — 

AT. Ta ^ 

^T=-js^-^^ 

In this formula 

A T = change in temperature. 
T = mean absolute temperature. 
J = the mechanical equivalent of heat. 
a — coefficient of expansion of material. 
d = density of material. 
S = specific heat of material. 
A 2? = change of stress. 
In centigrade units for a temperature of about 20° C. this 
formula gives for steel 

A T = - -000012 A p ■ 

Corresponding to a stress change of 20,000 lbs. per sq. in., 
therefore, we have a temperature change of only '24° C. 
The extreme delicacy of the method makes it suitable for 
use only under circumstances in which great care is taken 
to exclude draughts. 

This subject is dealt with in detail by Professor Coker in his 
Cantor Lectures before the Society of Arts 1913. In this 
paper the results of thermal tests upon a channel section were 
given, and there was a marked departure from the straight- 
line relation in the stress diagram, this being accounted for by 
the asymmetry of the section. Other experimental resuhs 
were quoted which showed that the tensile yield point as 
determined by the thermal method agree very closely with that 



THE TESTING OF MATERIALS 411 

found by extensometer. In the curve of temperature plotted 
against load, there is a sharp cusp at the yield point because 
the temperature then rapidly rises instead of falling. In 
these experiments, the observed results have to be corrected 
for cooling eSects . Although of very great interest, the method 
seems rather too delicate for very extended application. 

The Optical Method of Testing Materials. — Sir David 
Brewster discovered a hundred years ago that when plane 
polarised light is sent through a piece of glass under stress, an 
effect is produced upon the light which is detected by the 
appearance of colour bands when viewed through an analyser. 
The optical aspects of the subject are beyond our present 
scope, but the reader will be able to study these from the 
bibliography given below. The mathematical problems in- 
volved were dealt with by Clerk Maxwell and others, and in 
recent years particular attention has been given to the applica- 
tion of the method to the. determination of stresses in various 
machine and structural details by Professors Alexander, Filon 
and Coker. 

If a beam of plane-polarised light is passed through a speci- 
men of transparent material, such as glass or xylonite, and 
Nicol's prisms in the polariser and analyser are set with their 
principal planes at right angles so as to cut off the light, no 
effect is produced if there is no stress in the material, but if 
there is any stress, a colour effect is produced, and regions of 
zero stress or equal and opposite principal stresses, such as the 
neutral axis of a beam, can be detected by a dark patch or 
line. If the material is elastic, the colour produced will be a 
measure of the difference of the principal stresses at any point. 
The stress corresponding to a given colour is determined 
numerically by experiment by uniformly loading a small 
specimen until the colour produced is the same as at any 
particular point of the model under consideration where the 
stress is required. In cases where one of the principal stresses 
is negligibly small, the stress thus obtained can be taken as 
equal to the greater principal stress required. 



412 



THE kSTRENGTH OF MATERIALS 



In Professor Coker's experiments very successful results 
have been obtained with models of various structional and 
machine details cut out of sheet xylonite. 

Fig. 195 * shows the results of his experiments upon a tie-bar 
eccentrically loaded ; the resulting curves of stress agree very 
well with the theoretrical straight-line variation, with the 




so 



Fig. 195. — Stresses in Eccentrically Loaded Tie-bar. 



exception of that at the highest load, in which the material 
begins to yield at the edges and the straight line bends over as 
shown. In this case the test-piece showed residual stress at 
this edge after the load had been removed. 

In experiments upon models of standard cement briquettes 
Professor Coker found t that the maximum stress was about 
1*75 times the mean stress (cf. p. 78). 

* Engineering, January 6, 1911. 
t Ibid., December 13, 1912. 



THE TESTING OF MATERIALS 



413 



Fig. 196 "^ shows the results of the experiments by the 
same investigator upon the distribution of stresses in 




Fig. 196. — Effect of Holes on Stress in Beams, 

rectangular beams with holes cut through them. In the 
upper specimen a hole is in the centre and in the lower one 
holes are formed half way between each edge and the neutral 
axis. In this case it wiU be noted that the maximum stress 
* Engineering, March 3, 1912. 



414 



THE STRENGTH OF MATERIALS 



does not occur at the outermost fibre, but at the outer edge 

of each hole. 

Professor Coker and Mr. W. A. Scoble, B.Sc.,* have also 

experimented upon the stresses in tie-bars with holes in them 

, ...,.., T-p width of plate 

such as occur m riveted lomts. it c = -,• -^ y ^ — r- 

"^ diameter of hole 

and f is the mean stress over the whole width of plate, they 

find that their results for one central hole may be expressed 

by the formula 

maximum stress _ 3 c 

mean stress c + 1 

Also if a is the radius of the hole, the longitudinal stress at 
a distance r from the centre of the hole on a normal section of 
the plate is expressed by the formula 



/, 



2+^:+ 



3 a* 



where p is the stress at a long distance from the hole ; while 
there is also a radial stress given by 






It will be noted that at the edge of the hole /,. = 3p, i.e. 
three times the stress some distance from the hole. 

The following results were obtained wdth a strij) 1 in. wide 
and '186 in. thick, the load being 100 lbs. 



Diameter of 

Central Hole 

(inches) 

1 
i 


Stresses in lbs. per sq. in. 


P 


mean 


, maximum 


549 
547 
568 
570 
613 


584 
620 
724 
868 
1035 


1470 
1560 
1770 
1850 
2040 



* Engineering, March 28, 1913; Society of Arts Journal, January 16, 
1914. 



THE TESTING OF MATERIALS 415 

Bibliography 

Brewster, Phil. Trans., 1816. 

Clerk Maxwell, Collected Pampers, Vol. I. 

P. Alexander, Phil. Soc, Glasgow, 1887.* 

KeTT,Phil.3Iag., 1888. 

Filon, Proc. Camb. Phil. Soc, Vols. 11, 12. 

Phil. Mag., 1912. 
Coker, Engineering, January 6, 1911; March 3, 1912; 
December 13, 1912; March 28, 1913. 
Journal Society of Arts, Cantor Lecture, 1913. 

* Reprinted in Alexander and Thomson's Elementary Applied 
Mechanics (Macmillan). 



CHAPTER XV 

FIXED AND CONTINUOUS BEAMS 

If the ends of a beam are fixed in a given direction so that 
they are not able to take up the inclination due to free bending, 
or if a beam rests on more than two supports, the B.M. and 
shear diagrams will be different from the cases of simply 
supported beams that we have considered up to the present. 

In the first case the beam is said to be iixed, built-in, or 
encastre, and in the second it is said to be continuous. 

We will consider how the shear and B.M. diagrams can be 
found for such beams, and will point out their advantages and 
disadvantages compared with simply supported beams. 

FIXED BEAMS 

If the ends of a beam are fixed in a horizontal direction, 
then the beam when bent takes up some form such as a b c 
(Fig. 197). If the ends were free it would assume the dotted 




Fig. 197. — Fixed Beams. 



form a' b &, and to get it back to the form a b c, negative 
bending moments, shown diagrammatically as due to forces 
Fj, Fg, have to be imposed upon it. The ends of the beams will 
therefore be subjected to bending moments which will be 
negative because they cause curvature in an opposite direction 

416 



PIXED AND CONTINUOUS BEAMS 417 

to that due to the load. This change in sign of the bending 
moment means that the tension and compression sides of the 
beam are reversed. We will consider the cases of fixed beams 
both from the graphical and the mathematical standpoint, as 
we did in the case of the deflections of beams. 

INVESTIGATION FROM GRAPHICAL STANDPOINT 

According to Mohr's Theorem, the deflected form of a beam 
is the same as that of an imaginary cable of the same span 
loaded with the bending moment curve of the beam, and 
subjected to a horizontal pull equal to the flexural rigidity 
(E I) . If the ends of a beam are fixed in a horizontal direction, 
the first and last links of the hnk polygon determining the 
elastic line will be parallel ; this means to say that the first and 
last points on the vector line on which the elemental areas of 
the bending movement curve are set down must coincide. 
But this is equivalent to saying that the total area of the 
bending moment curve for the fixed beam must be zero. 
This enables us to enunciate the following rule — 

// the ends of a beam are fixed in a horizontal direction at the 
same level, and the section of the beam is constant along its length, 
there will be negative bending moments induced, and the area of 
the negative bending moment diagram will be equal to that due 
to the load for the beam if considered freely supported. 

We will speak of the negative bending moment diagram 
as the " end B.M. diagram," and that for the beam freely 
supported as the " free B.M. diagram." 

The problem now divides itself into two cases : (a) That in 
which the loading is symmetrical. (6) That in which the 
loading is irregular or asymmetrical. 

Symmetrical Loading. — If the loading is symmetrical, 
then the beam looks the same from whichever side it is viewed, 
and so the end bending moments will be equal, and their 
value can be found by dividing the area of the free B.M. 
diagram by the span. This will be made more clear by 
considering the following cases — 

EE 



418 



THE STRENGTH OF :\L\T£RIALS 



(1) Uniform Load on Fixed Beam. — Let a uniform load 

of intensit}- w cover a span a b (Fig. 198) of length /. The free 

B.M. curve is in this case a parabola a c b, with maximum 

IV l~ 
ordinate -^• 

o 

Therefore, since the area of a parabola is two-thirds of the 




Fig. 198. — Fixed Beam with Uniform Load. 

area of the circumscribing rectangle, area of free B.M. curve 

2 , w P w P 
= 3' ''-J- --12 



End B.M. = 



wP 
12 



J 

IV P 



IV P 

12 



Then setting u^^ a e and b f equal to , ^ and joining e f we 

get the end B.M. diagram, and the effective B.M. curve is 
the difference as sho^Ti shaded. At the points G and h the 
B.M. is zero, and these points are called the points of contra- 
flexure, the curvature of the elastic line changing sign at these 
points. 



FIXED AND CONTINUOUS BEAMS 419 

Suppose the point g is at distance x from the centre of the 
beam, then the ordinate of the parabola must be equal to -,^ 

XjLI 

10 /P o\. w P 



1-e. 2 1^4 ^'j - 12 



w X- 



X' 



IV P 

24 

12 

I 



X = 



2V3 



Distance of g from e = ^ — x 



2 ^ 2 2V3 

= -211 Z 

Shear Diagram. — With symmetrical loading the shear 
diagram will be the same as for the simply supported beam. 
This is because the shear at any point of a beam is equal to 
the slope of the B.M. curve at that point, and the slope of the 
B.M. is not altered in the case of symmetrical loading because 
the base line of the diagram is merely shifted vertically. 

Deflection. — The deflection at the centre can be found as 
before by considering the stability of the imaginary cable a^ Bj^. 

Considering the stability of the left-hand half of the cable, 

then taking moments about a^, we have 

E I X 8 = P ^1 - P ^, 

= P (2/i - ^2) 
In this case P = area of one-half of the free B.M. curve, 





2 I wP wP 




~ 3 ■ 2 ■ 8 " ~ ^24^ 


Vi 


51 
~ 16 


2/2 


I 
~4 


EI X 8 


wl^ /5l l\ wP 

24" Ue "iJ ~ 384 


.-. 8 


wP WP 
~ 384EI "" 384 EI 



420 



THE STRENGTH OE MATERIALS 



It will be noted that this is one-J&fth of the deflection for a 
freely supported beam with the same loading. 

(2) Isolated Central Load on a Eixed Beam. — In this 



case the area of the free B.M. curve 



1, WJ 

2^ ^4 ~ 8 



.-. End B.M. = 



8 



Z = 



Wl 




Fig. 199. — Fixed Beam with Central Load. 

The B.M. and shear diagrams are as shown in Fig. 199, the 
points of contraflexure being at J and J span. 
Deflection. — As in the previous case we have 
E I X 8 - P (2/1 - 2/2) 
Wl I Wl^ 



In this case P = 



8 



16 



Ui 



EI.8 = 



8 = 



I 
3 

/ 

WP 



16 
Wl 
192 EI 



4 



WP 
192 



FIXED AND CONTINUOUS BEAMS 



421 



This is one-fourth of the deflection for a freely supported 
beam with the same loading. 

* Asymmetrical Loading. — In this case the end B.M.s 
will not be equal, and in this case, in addition to the condition 
that the areas of the end B.M. diagram and free B.M. diagram 
must be equal, we have the further condition that their centres 
of gravity must fall on the same vertical line. 

This can be proved as follows : Considering the imaginary 
cable and taking moments about one end, the tension at the 




Fig. 200. — General Case of Fixed Beams. 



other end passes through the point so that its moment is zero. 
Therefore the moment of the B.M. diagrams about this point 
must be zero. Since the areas of these diagrams are equal, 
their centres of gravity must be at the same distance from the 
given point. 

Let a span a b, Fig. 200, of length /, be subjected to any 
irregular load system which produces a free B.M. curve AcdB, 
and let the centre of gravity of that diagram lie upon the 
vertical line G G. Suppose the end B.M.s are M, and M,„ and 
A a and b b are set up equal to these end B.M.s, then the 
trapezium a a 6 b is the end B.M. diagram, and the conditions 
that have to be satisfied are that the area of the trapezium 



422 THE STRENGTH OF MATERIALS 

shall be equal to the area of the curve a c d b, and that its 
centre of gravit}^ shall lie upon the line g g. Join a b, thus 
dividing the trapezium into two triangles, and draw verticals 

X X and y y at distances equal to from a and b. The centres 

of gravity of the triangles a a b, b « 6 lie on the lines x x and 
Y Y respective^, and our problem resolves itself into dividing 
the total area of the curve a c fZ b (which area we will denote by 
a) into two areas acting down the lines x x and Y y. This is 
effected bj^ treating the areas as vertical forces, and setting 
down a vector line 0, 1, to represent the area a. Taking any 
convenient pole p, we then join p and 1 p and draw x g, g y 
across the verticals x x, g G, Y y parallel to p, 1 p respectivel}', 
and joinxy; then drawing p 2 parallel to xy, 1, 2 gives us 
the area which must act down the vertical y y, and 2, that 
down X X. 

Then M, x ^ = area of triangle a a b = 2, 
• M =2,0x2 

Similarly M„ = ' , 

This enables the B.M. diagram to be drawn. 

Shear Diagram. — In this case as the end B.M.s are not 
equal the shear diagram will not be the same as for a freely 
supported beam, but the base line will be shifted. Since the 
shear at any point is the slope of the B.M. curve, the base line 
of the shear curve will be shifted downwards by an amount 

" ' " " because this is the change in slope of the base line 

t 

of the B.M. diagram between the freely supported and the 

fixed beam. If in the figure ac e d b represents the shear 

diagram for a freely supported beam with the given loading, 

then the effect of building-in the ends of this diagram is to 

]M — M 
lower its base line by an amount a a' = b b' = ^ ' , ^ ", thus 

giving the diagram a' c e' d b'. 



FIXED AND CONTINUOUS BEAMS 423 

Special Gases. — 1. Fixed Beam with Uniformly In- 
creasing' Load. — Let a beam ab of span I be subjected 
to a load of uniformly increasing intensity, the intensity at 
unit distance from b being w tons per ft. run, the total load 
being W. Then, as shown on p. 139 for a freely supported 

W 2 W 

beam, R„ = ,^ , R^ = — and the free B.M. diagram is a para 

bola of the 3rd order, the maximum B.M. being equal to 
•128 WZ and occurring at a distance '577 I from b. Then 

the area of this free B.M. diagram is equal to -r^- and its 

centre of gravity occurs at a distance ^ from b. This can 
be proved mathematically as follows — 

Area of B.M. curve = / M d x 



/ 
'w P X 10 x^\ , 

"6 6 ) '^ ^ 

w P x'^ w x* 

;~I2 "2T + ^_ 

The area = when x =^ 0. . ' .- c = 

_wl^ _wl^ _wl^ _^P 
•■• ^^^^ - 12 24 " 24 ~ 12 



First mt. of B.M. curve about vertical through b = jM.xdx 


I 

fw P X^ IV X^\ -. 

V 6 ~ Q^ '^"^ 

I 

/ 

wP X^ w x^ 
f8 30 + ^\ 

Moment = when x = 0. .* . c^ = 

-r^. . , wl^ wP wl^ 

'. 1^ irst moment = ,„ — „^ = .^ 

18 30 4o 



424 THE STRENGTH OF IVIATERIALS 

1st moment 



. ■ . Distance of centroid from vertical throngrh b 



area 



wl^ 


wl^ 


45 


■ 24 


24 Z 


8Z 


45 


15 



This fixes the Hne g g, and the areas that must be considered 

W P W l^ 
as actmg up x x and y y respectively are thus ^ and -^ 

\YP 31 

since the total area -^^ acts at distance ^ „ from y y. 

. • . Taking moments about y Y we have- 

Area actmg down x x x 7^ = -tt. x , ^ 

6 iZ ID 

. ^. , WP I WP 

. • . Area actmg down x x = ^„ -^ v. = >./^ 

bi) 6 ZO 

„ WZ2 2 WZ 

M — V — = 

20 Z 10 
,, WZ2 2 Wl 

M — - V = 

30 Z 15 

The resulting B.M. diagram then comes as shown shaded in 
Fig. 201. 

The amount of shifting of the base line for shear will be 

wz wn w 7W 

^TT. — , K -^ Z = ^^ SO that the shears at the ends are - ^ „ 
_ 10 15 J SO 10 

3 W 

and ^ respectively, the shear curve for the fixed beam then 

coming as shown. 

2. Non-central Isolated Load. — The following construc- 
tiori can be used for this case : Let a b, Fig. 202, represent 
a fixed beam of span Z carrying an isolated load W at a point 
p, the distances of which from a and b are a and b respectively. 
First draw the "free bending moment" diagram adb, 

i. e. set up p D = — , — , and join d to a and B. Project d 

horizontally to meet the vertical through the support b at 
E and join e a. 



FIXED AND CONTINUOUS BEAMS 425 

Then pf = Reverse or end bending moment at b = M^; 
and F D = reverse or end bending moment at A = M,. 




Fig. 201. — Fixed Beam with Uniformly Increasing Load. 

Therefore set up b h = r p and a G^ = f d and join g^ h, 
the complete bending moment diagram then coming out as 
shown shaded. This may be done by projecting f horizontally 
and drawing d g^ parallel to F a. 



420 



THE STRENGTH OF MATERIALS 



Proof. — To prove this construction ^\c must obtain values 
for the reverse bending moments for this case. Referring to 
Fig. 203, it will be remembered that in fixed beams the con- 

Jl ./J 




Fig. 202. — ^Fixed Beam with Isolated Load. 

ditions to be satisfied are that the " free '' and " end " bending 
moment diagrams a b d and a G^ H b respectively must be 
equal and op^iosite in area, and must have their centroids upon 
the same vertical line. 




Fig. 203. — ^Fixed Beam with Isolated Load. 

The first condition gives us 

—^^^ — — - = area of A a d b = i a b . r ]) 
I .Wab Wab 



1 



2 



• (1) 



FIXED AND CONTINUOUS BEAMS 427 

The end bending moment diagram a Gj. h b may be con- 
sidered as divided up into two triangles a g^ B, b g^ h, whose 
centroids act in the " third Unes " x x and y y respectively. 

We have next to calculate the position of the centroid g of 
the free bending moment diagram. According to the ordinary 
rule, the centroid g will be one-third of the way up the median 
line c D. 

1 1 fl, 

•*• ^ " 6 ^^ ~6 3V2 ""y ~3 

Regarding the areas of the triangles ag^b, bg^h, adb 
as concentrated in the lines x x, y y, and g g respectively, 
we have by taking moments about the line x x 

Area ofAADB xa; = area of A b g^ h x ^ 

o 

i. e. from (1), — ^ — . .t = | M,, ? x ^ 

M,J2 ^Wab _Wab a 
■ • 6 ~ 2 '^ ~ '2' 3 

_ w an 

~ 6 
• M = ^^^ 

• • -^'-Ljj 72 

Similarly, M^ = „ — 

As a check (M, + M„) = '^f ^ + ^ |^ 

Wa6. ,, Wab , Wab 
= — ^ (a 4- 6) = ^2 '^= I 

(M, -f MJ I Wab 



(as in (I)). 



Wab 
Now, smce — -. — = p d 

„ _PD X a. PD X 6 



428 THE STRENGTH OF MATERIALS 



but in Fig. 202, by similar As 



P F 


AP 


a 


B E 


AB 


~ r 


P F 


B E 
I 


.a p D . a 



Similarly, e h = d f = — , " = M, 

Position of Load for !MAxiMr^r Reverse Bending 
Moment.- — The position of the load for a maximum value of 
the reverse or the end bending moment M, is obtained by 

putting -^' = and noting that a = {I — b), 

dh " - ^ 
W d{ah^) _ 

i^ dh 

i.e. ^fP = 

do 
2 6 Z - 3 62 = 

I. e, = o • or 

Taking the first value, which is clearly the maximum, then 



M _^3V3y _4WJ _ W| 
P ~ 27 ~ 6-75 

Therefore the maximum reverse bending moment for an 

Wl 

isolated load is equal to w;^^, and occurs when the load is at 

one-third of the span. 

!MAXoir>E Positive or Intermediate Bending Moment. — 
Referring to Fig. 202, the maximum intermediate bendmg 
moment occurs at the load point p and is equal to D J. 
. • . Maximum intermediate bending moment 
= !Mp = D J = P D — J p 
Wab 



FIXED AND CONTINUOUS BEAMS 429 

Now, remembering that {a + b) = I 

JP=PF + FJ 

= M3 + (M.-MjJ 



= M„.'; + ^^* 

_ Wtt^6 a W.ab'b 
= Waft ^^, _^ j,^ 

~ ^ \ Z2 

= — y^ — .2 ah 
_ 2 . W g^ 6^ 

To get the maximum vahie of this for any position of the 
load put ", - = 

'• '• ~l^' \ d^ / " ^ 

d.a^a^ -2al -\- a") 

I.e. J = 

da 

i.e. 2a?-^ - 6ft^Z + 4a3 _ 

i. c. Z^ ~ 3 a ^ + 2 a'-^ - 

(^ - a) (Z - 2a) = 

I . 
I.e. a — ^, or I. 

Taking the former value, which is the only one possible, wc 
have 

Therefore the maximum intermediate bending moment 

W I 
occurs when the load is at the centre and is equal to -^— . 

o 



430 THE STRENGTH OF MATERIALS 

Graphical Method of finding g g. — If the nature of the 
loading is such that the position of the hne g g cannot be 
calculated without difficulty we may proceed as follows : 
Divide the free B.M. diagram acb up into a number of 
vertical strips, not necessarily equal, and draw vertical force 
lines through the centres of these strips and set down the 
ordinates on a vector line, and wdth any pole draw a link 
polygon. The point Avhere the first and last links meet will 
be a point on the line G G. This is the same method as 
adopted in finding the centroid of a figure by Mohr's method 
(Chap. LX). The area of the B.M. diagram can be found 
by sum-curve construction, and the problem completed as 
indicated with reference to Fig. 200. 

INVESTIGATION FROM MATHEMATICAL STANDPOINT 

As we have previously seen 

r M 

Slope of beam = / -^ ^ ^ ^ 

If the end of the beam is built-in this slope must come zero 
at the two ends. 

Consider the following special cases — 

(1) Uniform Load on Fixed Beam. — Taking the in- 
tensity of load as w and the centre of the beam as origin, theh 
considering a point at distance x from the centre, for the 
freely supported beam we have 

M,. = |(^' - x^) (See p. 266.) 

Let the effect of the building-in be to cause an end B.M. = M^ 

Then for the fixed beam M., = ^ (4 " ^') " ^^ 



Slope -= / -p, -J- d X 



1 (w P" X w x"^ 



EIV 8 6 

Slope is when x = 0. . • . c = 0. 

Also slope must be when ^ = 9 



- M,, X -f c j 



FIXED AND CONTINUOUS BEAMS 431 





.-. ' 




48 


i'" 






I 


'. c. M, X 2 


wP 
16 

w P 

~ 24" 
wP 
~~ 12 


wP 
48 


• 






To obtain the deflection we integrate again, 


and we get 




deflection 


=/ 


/EI^^^ 












1 
~EI 


/i^; P x^ 
\ 16 


w x^ 
24 


M, a;2 

2 


+ Ci^ 






1 
"~EI 


Z!/; P x^ 
V 16 


w x^ 


w P x^ 
24 


+ Ci^ 






1 
~ EI 


/i/; P x^ 
V 48 


w x^ \ 
24 +^V 




This is 


when x 


I 

~ 2 












•*• 


192 


wP 
384 + ^1 

~ 384"" 


= 




^ 



.'. When .T = 0, deflection = =ri\ 

E I 

"" 384 E I 

.-. Maximum deflection ^ 3^-^-^ ^ ^ 3^^^ ^ 

The B.M. and shear diagrams are then as shown on Fig. 198. 
(2) Isolated Central Load. — Taking as before a span I 
and the centre as the origin, if the load is W, for a freely 
supported beam we have 

^■'" "" 2 \2— ^. 
.'.If the end B.M. due to fixing the ends is M,, we have 
for the fixed beam 

M. = ^ (2^ - a;W M. 



432 THE STRENGTH OF IMATERIALS 

/M 

When X ^ O,«slope = 0. . • . Cg = 0. 
When X = x, slope also =- in this case. 



He have : = -^ -~ M, ^ 



V 8 16 ■' 2/ "" E I 



X 



I W/2 

M . - =: - 

' 2 16 

M ^^^ 

8 



To get the deflection we integrate again, then 
deflection = / / t^i j d x 

1 /Wlx^ Wx^ M. a;2 



This is when x = ^ 



EIV 8 12 2 

_ J^/Wl .t2 _ W^ 
~EI\ 16 12 

I 



+ c, 



+ C3 



WJ3 _ WP - 

64 96 + ^3 - ^ 

' ^3 ^- 192 

c. 



When X = 0, deflection -^ --t 

E I 

W /■■ 

.". Maximum deflection = 



192 E I 



* (3) Fixed Beam with Uniformly Increasing Load. — 
Let a span a b of length I have a uniformly increasing load, of 
zero intensity at the point B, and let the intensity of load at 
iHiit distance from B be i:; units per ft. run. Then taking the 
end B as origin, we have in the case of the freely supported 
beam 

,^ wl'^x w x^ 

M'= 6—6 



FIXED AND CONTINUOUS BEAMS 433 

Now let M, and M„ be the end B.M.s, then the negative 
B.M. at distance x from b is equal to 

. * . for the fixed beam 

/M 
=Frj • d X 

_ 1 I'l^Z^a;^ wx^ /M^j-MAa;2 ^ 

~ EI 1 12 ~ 24 ~ ^-^"^ "" V I J 2 + ^^/ • • ^^' 

When X = 0, slope = 0. . • . c^ = 0. 
Also when x = l, slope = 

• • 12 24 ^^^^'^~ 2Z ~^ 

2 ■ 2' ~ 24 

.-. M. +M3 = '^j2' (^) 

To get another relation between M, and M^, consider the 
deflection ; 

then deflection = / / =nrj d x 

_ 1 \wl^^^ wx^ M,x^ ( M, - M, \ ^ , 1 .ON 
~EI\ 36 120" 2 V I y-e"^^^/ "^^^ 

Deflection = when x = 0. . '. C5 = 0. 
Also deflection = when a; = Z 



36 120 2 V Z / 6 



- M., Z^ _ M, j2 M„J2 ^ ^^^Z5 _ w; Z5 
2 6 ^ 6 ~ 120 36 

M, Z2 M, j2 ^ ^wl^ 
•'• 6 "^ 3 ""360 

.-. M. +2M,= ^^^^ (4) 



FF 



434 THE STRENGTH OF MATERIALS 

. • . Combining (3) and (4) Ave get 

^ _ 1 W P IV P 

" "~ "60^ 12 

~ 30 ~ 15 

• ' ^ ^- 12 30 ~ 10 

The B.M. diagram then comes as shown in Fig. 201. In all 
the above cases we have assumed that the beam is of constant 
cross section along its length. If such is not the case, the end 
B.M.s can be found by taking the corrected B.M. diagram 
as explained in the chapter on the deflections of beams. 

Advantages and Disadvantages of Fixed Beams. — 
We have seen that, in the examples that have been considered, 
a fixed beam is stronger than the corresponding freely sup- 
ported beam, and that the fixed beam has smaller deflections 
and is thus more rigid. In most cases, moreover, the maximum 
B.M. occurs at the abutments, Avhere the beam can be strength- 
ened -wdthout adding materially to the bending moments and 
thus increasing the stresses. In the freely supported beam, 
on the other hand, the maximum B.M. occurs at the centre. 
Avhere an addition of weight to strengthen the section would 
add materially to the B.M. The reason Avhy such beams are 
not more commonly adopted is because, in fixing in the ends 
securely, the tangents at each end to the beam must be 
absolutely horizontal, and any deviation from this will alter 
the stresses, and any difference of level at the two ends due 
to unequal settlement would cause considerable stresses in 
the beam. There is also considerable stress due to change in 
temperature if the beam is securely built-in to the masonry, 
and all these points make the actual stresses in any practical 
case somewhat uncertain, so that many designers do not use 
this type of beam. All the above objections can be obviated 
by cutting the beam through at the points of contraflexure 
and resting the centre portion on the two end portions. This 
is the principle of the cantilever girder construction and for 



FIXED AND CONTINUOUS BEAMS 



435 



large spans is very economical. This is shown diagrammatic - 
ally in Fig. 204, in which a fixed beam a b is shown divided 
at the points of inflexion c and d and the centre portion is 
represented as hanging from the end portions. The B.M. in 
the centre portion will be the same as for a freely supported 
beam of span I loaded in the given manner. The B.M. for 
the cantilever portions will be the same as for cantilevers of 
span Zj loaded with the given loading and also with loads at 
the ends equal to the reactions at the ends of the centre 
portions. In the figure, uniform loading is shown, and in 

such case these reactions are each equal to -^ . 



It will be 




Fig. 204. 

found that the resulting B.M. and shear curves obtained in 
this way will be the same as shown in Fig. 198. The deflec- 
tions can also be found by adding together the deflections at 
the centre of the centre portion and at the end of one of the 
cantilever portions. 

Fixed Beam with Ends not at same Level. — Suppose 
that a fixed beam a b. Fig. 205, has its ends at a different 
level, then apart from the loading on the beam, the deflected 
form of the beam will be as shown in the figure, the point of 
contraflexure being at the centre point c. 

The deflection 6 e of the portion a c, assuming the beam 
divided at c, will be equivalent to that due to a weight P 
hanging downwards at c, but for a cantilever with load at end 

WZ3 



8 = 



3 EI 



436 THE STRENGTH OF MATERIALS 

In this case we have 



eh = 
. P - 



Z\3 



3EI 
24 EI X eb 

P 
12^1 X cf 

12EI X <Z 





End B lourer End /^ lou/er. 

Fig. 205. — Beams with Ends fixed at different Levels. 

The B.M. diagram due to this is a triangle c a^ d, a^ d being 
equal to P x ^- 



.*. Ai D = 



12EIxc^ 6EIcZ 



11 



Z2 



Similarly the portion c B is as if it had a load P at its end 
acting upward, the B.M. diagram for this portion being c b^ e, 
Bj E being equal to a^ d. 

Therefore, this diagram must be combined wdth the ordinary 
diagram for a fixed beam if the ends are at different levels, the 



FIXED AND CONTINUOUS BEAMS 437 

figure showing the effects for the case in which b is lower than 
A, and also that in which a is lower than b. 

The condition that the end B.M. diagram must be equal in 
area to the free B.M. diagram still holds in this case, but their 
centroids are not on the same vertical line because there is a 
resultant deflection at one end. 

It can be shown by considering the stability of the imaginary 

cable of Mohr's Theorem, that E I x (Z = area of B.M. 

curve X horizontal distance (g) between the centroids of the 

free and end B.M. curves. 

u) Z 
i.e.'El X d = -^- X I X g 

_ 12 E I X <Z 

Now, if M^ and M^ are the end B.M.s, the end B.M. 
diagram is a trapezium. 

I Z /2 M3 + M, 



• * • ^ 2 3 V M3 + M, 
_ I (M, - M,) 

6Tm;+mj 

_ 12 E I«^ 

Now, in the figure M^ — M^ = 2 a^ d 
M, - M„ 6 E I ^ 



A, D 



P 



This gives the same result as the previous reasoning. 

Beams with Cleat Connections, etc. — In building work 
the girders are usually connected to the stanchions or columns 
by means of cleat connections, which, owing to their rigidity, 
make it doubtful whether the girder will act as a freely sup- 
ported beam, although their strength is almost invariably 
calculated as such. Neither is an ordinary cleat sufficiently 
rigid for the girders to be considered as fixed at their ends. 
The actual B.M. diagram for such beams will be somewhere 



438 



THE STRENGTH OF MATERIALS 



between that for a freely supported beam and a fixed beam. 
It has been suggested that these beams should be treated as 
" half fixed," that is, that the end B.M.s should, in the case 

of uniform loading, be taken as ^ . The B.M. diagram then 

comes as shown in Fig. 206. It will be noted that the maxi- 

w P- 
mum B.M. in this case is still ^ as in the fixed beam, but 

such B.M. now occurs in the centre. 




t Fig. 206. 

In beams where the tensile and compressive strengths of the 
material are different, as in cast iron and reinforced concrete 
beams, it must be carefully remembered that at the ends the 
tension side is at the top, and so the additional strength must 
be placed at the top at these ends. 

It must also be carefully remembered that in all the cases 
we have assumed that the cross section of the beam is constant 
along its length, and the results obtained will iiot be true if 
such is not the case. 



CONTINUOUS BEAMS 
If a beam is continuous over a number of supports a, b, c, 
the deflected form of the beam has to take some shape such as 

a 



^ 




Fig. 207. 



shown in Fig. 207, the curvature changing in direction at the 
points a, h, c, d. As in the case of fixed beams, this change in 



FIXED AND CONTINUOUS BEAMS 



439 



the curvature means that a negative bending moment occurs 
at the supports, such bending moment being called in future 
the " support B.M." 

Consider first the case of a continuous girder, a, b, c, Fig. 208, 
of two equal spans, each of length I, subjected to a uniform 



cr)crTTTyrrrrrrtrirnrcnrn_ 




^url 



B.M.Diaarcim on 'strai(^hf oase 
Fir;. 208. — Uniformly Loaded Continuous Beam of two equal Spans. . 

load of w tons per foot run, the supports a, B, and c being on 
the same level, and the beam being of uniform cross section. 
Now imagine the centre support removed, then there would be 
a central deflection 8, given by 

5 w; (2 Z)* 
~ 384 E I 

Now, if the centre support be replaced, the pressure R„ on 



we see that R^ = R^. = 



440 THE STRENGTH OF IVIATERIALS 

it must be such that as a central load it causes an upward 
deflection equal to 8 

R X (2 Z)3 
48 EI 
n X {2lf _5w{2 /)* 
• ■ 48 E I ~ 384 E I 

®= 8 =4 

5 W 
or if W is the load on one span, R = - t— 

. • . Since R^ = R,. from sj^mmetry, and R , + R,, -f R, = 2 W, 

3 W _ 3 jW 

" " 8 

W 

In the ordinary case of two separate spans R, = R^ = -^ 

. • . Support B.M. diagram wiU be as if there were an upward 

W 
force of ^ acting at a and c. This causes at b a B.M. 

= -^ X t = so that the negative B.M. at b = ^ = 

o o o o 

and the B.M. diagram for the continuous beam then comes 
as shown in Fig. 208. 

As the reactions are ^wl Sit a and c, the shear diagram will 
have an ordinate equal to 1 1^ Z at these points ; the shear then 
decreases uniformly from c to b until it has a value — ^wl 
at B. It then increases to -j- ^wl, since 'R^ = fwl, and then 
decreases to — ^wl again at a, the shear diagram then coming 
as shown in the figure. 

The points of contraflexure g, h, where the B.M. is zero, 

occur at distances . from b. 
4 

This can be shown as follows — 

Let H be at distance x from c. 

Then negative B.M. due to support B.M. = = "o 

o o 

•J/1 / cr ID 'X, 
positive B.M. for freely supported beam = — ^ ^ 



FIXED AND CONTINUOUS BEAMS 441 



These must be 


equal, so 


that 








w Ix 


wlx 


to x^ 




8 


~ "2" 




2 




X 


I 


I 


3Z 




• • 2 


~ 2 


8 


"" 8' 




.*. X 


3Z 
4 







. • . distance from b = I 1- = t 

4 4 

If the B.M. diagram be reduced to a horizontal base, the 

lower diagram shown on the figure will be obtained. 

The maximum intermediate B.M.s will occur at distances 

3 I 
- from c and a. 
8 

^, .„ , , , wl 31 w /31Y wl 31 

They will be equal ^^ -j- - ^ ~ 2 ' \Y) ~ "8~ ' 8 

72 ^1 _ A _ 1 
^ Vl6 128 64 

^ 9wJ^ ^ 9WZ 
" 128 ~ 128 

* Two Equal Uniformly Loaded Spans with Sup- 
ports not on same Level. — Now consider the case in which 
the centre support B is at different level from a and c, and let 
B be at distance h below A c (Fig. 209). 

As before, if the support b is removed, there will be a central 

deflection S = 004^ Vpj" 

The reaction at b is now only sufficient to cause an upward 
deflection equal to 8 — ^. 



3 



K, (2 I) 
^ ~ 48E I 



^' ^ irtf ^^ ~ ^^ 

_ 48_EI / _h 
(2Q3 ^V 8 



442 THE STRENGTH OF MATERIALS 

^ _48EI 5m;J2j)*/ h 
' ' -^" {2 If ^ 384 EI V 8 

= 4 (1- J (1) 

_ 5wl 5 h w I 
~ T 4T~ 

•■• ^'^^^■•= 8+ 88 

^"^ ^-^/l_") (2) 



2 8 V 8 

.•. Reasoning as before, negative B.M. at b dne to the 
second portion of R^ or R, 

_ If Z^ / 5h 

^.e.M„=.-/(l-") (^) 

.'. the B.M. curve will be somewhat as shown shaded, the 
position of d depending on the value of - 

Now consider the following special values of /*. 
If h = 0, M,, = as in the previous case. 

o 

If j^ = M„ = 0, and the B.M. diagram is the same as for 
o 

two simply supported beams. 

If h = S, M., = ''J'- (1-5)= -J'' 

This is the same as we should have obtained for a simjoly 
supported beam of span 2 /. 

Now let h = ^ 
5 

i/j 72 1*1 72 

Then M„ = „ (1 + 3) = ^ . This is the same as if the 

supports A and c were removed and the beam were two 

cantilevers b a and b c. The free deflection at the ends is 

wl^ 3 

then = o -c^ T5 ^^^ ^^is ^ill ^® found to be equal to - 8. 



FIXED AND CONTINUOUS BEAMS 

38 



Now h must lie between 8 and 



443 

for the beam to act as 



a continuous beam, therefore take points e, e' on the vertical 




Fig. 209. — Continuous Beam with three Supports 
not on same Level, 

through B, such that b e' = b e = ^ , then the closing line 

of the B.M. diagram for the continuous beam with the supports 
at different levels must lie between a e c and a e' c. 
The following example on this problem is interesting — 
A continuous beam of uniform section and two equal spans I 
has a uniform load of intensity w, and the supports a b c are 



444 THE STRENGTH OF MATERL4LS 

initially level. The support columns are, however, equally 
elastic, the force necessary to cause V7iit compression being e. 

Find the central reaction and B.JI. 

5 IV (2 /)"* 
If the centre column is removed, 3 = ^o .V. 4- 

384 E i 

T> /9 7x3 

The upward deflection due to R^, = S^ = ~vo ^i^-~ 

Then 8 — 8^ = difference in level between final posijtions of 
A, B, and c. Now let R3 = if Z -f 2 /, 2 / being the additional 
reaction due to the beam being continuous, then 

R. = R.- = ^' - / 

. • . Sink of central column = 



Sink of end columns 



e 
w I 

'2 - ' 



e 
Difference = S — S^ = ( ^^ -f 3 / 



1 /3 R3 ; 

f-ivl 



1 /S R« 

~ — tv 



e 



; ? ) = S - 8^ 



R. 



_ 5w]^ _ R„ P 
~ 24 E I 6 EI 

P 3 \ DWl^ . wl 



6EI'2e/ 24 EI' e 

5 IV l^ . IV I 

.' . R« = 



24 EI e 



J3 _^ 3 

6EI ' 2e 



= IV I 



24 EI e 
I P 
I 6 E 

5 . 6EI 



= IV I \ 



I ' 2eJ 

[5 , 6En 
J 4 ' eP I 

I - ^ M 



FIXED AND CONTINUOUS BEAMS 445 

Reasoning as before, we then get 



M„= 2 



5 6E^ 

4~^ eP 



1 + 



9EI 



- 1 




It will of course be noted that if the piers had been of the 
same material and of areas proportional to the reactions, the 
amount of sinking due to their elasticity would have been 
equal, and the B.M. diagram therefore would remain as shown 
in Fig. 208. 

* The Theorem of Three Moments. — We will now find 
the relation which must exist between the support bending 
moments and the loading for a continuous beam of any number 
of spans, the supports all being on the same level. 

Let A B and b c be any two consecutive spans of length l^ 
and ^2 of a continuous beam of any number of spans, and let 
A e B, B / c (Fig. 210) be the free B.M. diagrams for the loading 
on these spans. Let G^ and G2 be the centroids of these free 
B.M. diagrams, and let them be at distances y-^, y<^ respectively 
from A and c, the areas of the diagrams being respectively 
Si and Sg. Then, if M^, M3, M^ are the support moments at 
A, B, and c respectively, Cla'peyroni' s Theorem of Three Moments 
states that 

M, ?i + 2 M3 {h + h) + MJ2 = 6 1 ^y^ + ^f^) 

We can prove this with the aid of Mohr's Theorem* as follows : 
Let a! b' c' be the deflected form or elastic line of the beam, 
then if the beam is of the same material throughout, and of 
constant cross section, the elastic line is of the same shape as 
that of an imaginary cable loaded with the B.M. diagrams and 
subjected to a horizontal pull equal to E X I. Now the 

* See p. 252. 



446 



THE STREXGTH OF >L\TERIALS 



tangent to the imaginary cable is common at the point b'. 
Let such tangent be at angle 6 to the line a' b', and let the 
perpendicular from a' on to it be of length ii\, the tension in 
such cable at b' being T„ ; then considering the stability of 
the imaginary cable we have by taking the span a b and taking 
moments round a' 

Tb' ^ Vi = moment of B.M. diagram about a' 

= Sji/i — ■ moment of support B.M. diagram about a' 



Mc 





e 




.^ 


h 


< 


K 


Ma 






^^ 


f^ 


^p:'^=-^-^ 


Pn 


/I 


\ 




B 






c 




'-^'^ 


/ 






7 


*-/2^ 






h 






6^ 




^ 






^.B' 


% 

^--*^ ■- 






»^ 






£-/o, 


^^^"^ 




r//c Line 



c 



Fig. 210. — Theorem of Three Moments. 



= Si2/i-M,^'x J 



Ar / - / - 



^*^«2 3 



(1) 



because the support B.M. diagram can be di\-ided with two 
triangles of area ^-^ and ^— ^^, the distances of their centroids 

from a' being respectively -^ and ^ \ Xow p^ = I^ sin 0, and 



T - 



EI 

cos 



, E I being the horizontal pull in the cable. 



FIXED AND CONTINUOUS BEAMS 44: 

m E I Zi sm 

• , T„, X ;Pi = ^-. = E I Zi tan ^ 



cos 



.•.EItan^=%i^-^A_^i (2) 

Now by considering the second span, as is the same for 
both spans and E I is constant, we get 

EItan^ = -(S^f -^^'^-2M^) (3) 

The — sign is used because the moments are taken in 
opposite directions. 

Then, combining equations (2) and (3) we get 



?! 6 6 V ^2 ^ ^ 

or M, Zi + 2 M, {I, + Z2) + M, l, = 6 (^"^-^^^^ + ^f ^^) • • • • (4) 

TA^5 z<s the general formula applicable for all loadings. 
If the loading is uniform over each sj)an and of different 
intensities Wi and W2, we get 

c _ ^ 7 w^l^^ _ IL\ l^ 



Similarly 



^1-2 



3 1 • 8 12 

2 



. • . In this case we have 
M, /i + 2 M, {l^ + I,) + M, I, = ^ {w^ l^^ + 2^'2 ^2') . . • • (5) 

If the load is of the same intensity w on the two spans we get 
M. Z, + 2 M„ (Zi_+ I,) + M, Zo ^ ^ {I,' + Z^-^) (6) 



448 THE STRENGTH OF MATERIALS 

* Reactions and Shear Diagrams. — As in the case of 
fixed beams, the shear diagrams for continuous beams will 
have their base Unes shifted, due to the change in slope of the 
B.M. curve. 

Consider any support, say b, and let r^ be the reaction at b 
due to the span ^^ if the separate spans were simply supported, 
Ri being the corresponding quantity for the continuous beam. 

Then change in slope of B.M. curve = " T" ' 

T5 , M„- M, 

.-. Ri = ri+-^^^ ^ 

Similarly if r^, R2 are corresponding quantities for the 
span Zg 

. , M, -M. 

^2 

. ■ . Total reaction at 

Then R^^ and Rg give the ordinates of the shear diagrams on 
either side of b. This will be made clearer in the following 
numerical example — 

A continuous girder, a b c d [Fig. 211), consists of three spans, 
20, 10 and 15 ft. long, and the first span carries 20 tons, the 
second 15 tons, and the third 10 tons, uniformly distributed. 
Draw the BM. and shear diagrams. 

First draw the B.M. diagrams as if the separate spans were 
freely supported. 

Now take the first two spans, then by the theorem of three 
moments 

M. X 20 + 2 M, X 30 + M, X 10 = ^ j^^ . 20^ + J^ . lo4 

But the end a is freely supported. . • . M, = 

. • . We get 60 M, + 10 M, = ^^' (s + J^) 

or 6 M„ + M, = 237-5 (1) 

Now consider the next two spans. Then we have 

M, X 10 + 2 M, X 25 + M, X 15 = ^ | J^ . 10^ + J^ . 15^1 



FIXED AND CONTINUOUS BEAMS 



449 



The end d being freely supported, we have 
10 M, + 50 M, = ^^ |l2 + isj 

or M„ + 5 M, = 93-75 (2) 

Solving the two simultaneous equations (1) and (2) we get 

M, = 37-75 
M, = 11-20 
.'. Putting up these values we get the B.M. diagram as 
shown in the figure. 



2.0 Tons 



B 



ao 



1 5 Tons 



/O' 




3. M. Diagram (Figs. 




lO Tons 



D 



15 



Ft Tons) 



'i-a\ 



Shear UiaciramfFiq'^ inlons) 
Fig. 211. — Continuous Beam of Three Spans. 



To get the shear diagram we first calculate the reactions as 
follows — 

M„ _ 20 _ 37-75 
" 2 20 



K 



2 ^ li 



8-11 tons 



Rb ==5^ + -™ + ^1 + ^^^ == 11-89 + 10-15 = 22-04 tons 



^ 15 26-55 , 10 , 11-20 , o^ , .. -. 
^^ = 2 - -10" + 2 + 15" = ^'^^ + ^'^^ 



K = 



10 _ 11-20 

2 15 



= 10-59 tons 
= 4-26 tons 



Total 



45-00 tons 



GG 



450 THE STRENGTH OF MATERIALS 

The shear diagram then comes as shown in the figure, the 
continuity of the beam altering only the base lines, and not 
the form of the curves. 

If there are more than three spans, consecutive spans are 
taken two together, and a series of equations obtained by the 
theorem of three moments. Further numerical examples will 
be found at the end of the chapter. 

* Continuous Beams with Fixed Ends. — If the end 
of a continuous beam is fixed, the end B.M. is obtained by 
imagining a beam to exist beyond the fixed end of the same 

c_ g 

f7 21- 

r 2 
o i 

S 3 8 ^ 

O IQ 75 O 

/C /O lO lO 10 /o 

^ J- J^ -3l. O 

j£ -£ ^ ^ O 

33 ££ 3S 38 

f7s JIJzS 7I¥q TfYZI A^Ui ^ 
O /o^ /o'f- ro4- fo^ 704- f 

f?7 ^\is ZT^I ^Tj3 ~I[^ ^^63 ^^ 

Jok- fo4- /o4^ /of- /^ /of I04 /a4 fof- /(H /of- /o4~ 

FiG. 212. — B.M. and Reactions on Uniformly Loaded Continuous 
Beams of Equal Spans. 

length, and loaded in the same manner as the last beam. This 
is because the fixing of ends makes the beam horizontal at 
such ends, and this occurs at the centre of a continuous beam 
symmetrically loaded. An example of this will be found in 
the worked examples at the end of the chapter. 

Equal Spans with constant Uniform Load. — In prac- 
tice the spans (Z) are often equal, and the uniform load {w) 
per foot run constant, the extreme ends being freely sup- 
ported. A diagram is shown in Fig. 212, from which the 
support B.M.s and reactions can readily be obtained for any 
number of spans up to six. 

Above the span Hnes are the support moment coefficients, 
which have to be multiplied by w P. 



FIXED AND CONTimJOUS BEAMS 451 

Below the span lines are the reaction coefficients, which have 
to be multiplied by w I. 

From these the B.M. and shear diagrams can be readily 
drawn. The student should check these by working them 
through by means of the theorem of three moments. 



* GRAPHICAL TREATMENT OF CONTINUOUS BEAMS 

In dealing with a considerable number of spans with 
irregular loading, the application of the theorem of three 
moments becomes a somewhat laborious process. Although 
the following general graphical method is somewhat involved 
and takes considerable time to explain, it is interesting and 
useful, and shows to what extent the graphical method of 
reasoning can be pursued. 

Consider the imaginary cable of Mohr's Theorem which 
gives the elastic line of a beam. It is a link polygon for 
the bending moments, drawn with a polar distance equal 
to E X I. 

Now the slope and position of the first and last links of a link 
polygon are quite independent of the exact distribution of the 
forces, provided that they have the same resultant in magnitude 
and direction. 

As we shall see later, we shall be able to obtain the support 
moments if we know the support tangents to the elastic line. 
Let A B (Fig. 213) represent a span of length ? of a continuous 
beam, and let A c B represent the free B.M. curve for the 
loading on it, A a and b b being the support moments, M^ and 
Mj,. If the centroid of the curve a c b is g, then the vertical 
G G is called the centroid vertical, and if the support B.M. curve 
be divided into two triangles a a b and b a 6, the areas of such 
triangles act down the right and left hand third lines x x and 
Y Y. Now replace the actual B.M. curve for purposes of 
finding the elastic line by single forces acting down and up 
the lines g G, x x, Y Y. 



452 



THE STRENGTH OF IVIATERIALS 



On a vector line, 

set down 1,2= area of free B.M. curve a c b = S 
,, ,, 0, 1 = area of triangle a a b = S^ 

>5 jj 2, 3 = ,, ,, abB = S„ 

Then with pole p at polar distance (/;) = E I if A^d is drawn 
parallel toOT,dh to If, hg to 2 f and ^ b^ to 3 p, cZ h and h g 
are called the mid links, and a^ d and g b^ give the support 
tangents. 




Fig. 213. — Continuous Beams — Graphical Treatment. 

Now in our problem Ave do not know the position of the 
points and 3, and we see that these would be known if the 
mid links were found, so that our j^roblem now reduces to 
that of finding these mid links. 

On both sides of the centroid vertical g g draw lines at 
distance p, and set down lengths l^, 2^ equal to 1, 2 and join 
them across, intersecting on the centroid vertical. These 
lines are called the cross lines. 

Now draw any vertical u u, then clearly the intercepts made 
by the vertical on the mid links and cross lines are equal. 



FIXED AND CONTINUOUS BEAMS * 453 

From this it follows that if a point on one mid link is known, 
a point vertically below it on the other mid link can be found. 
Again, let the right-hand mid link of this span meet the 
left-hand mid link of the next span in a point j, Fig. 214, on 
a vertical line q q. 




Fig. 214. — Continuous Beams — ^Fixed Points. 



They are similar 



Then consider the triangles g j k, f 2 3. 

j k _ x-^ 

p X j k = 2, 3 X X;^ 

= Xj^ X area 6 as 



454 THE STRENGTH OF MATERIALS 

Similarly considering the triangle on the other side of Q Q 
we should have 

Where Zg is the length of the next span. 

Further, x-^^ -\~ X2 = k (li + h) 

h 
• • ^1 ~ 3 

x<, - 3 

. • . Q Q is at a distance = -^ from y y, and is thus called an 

inverted third line. 

Determination of "Fixed Points." — Let ab c (Fig. 214) 
represent two consecutive spans of a continuous beam, and 
let the third lines be drawn as shown. 

Suppose that we know that the right-hand mid link of the 
span A B passes through a fixed point r. Let this mid link 
cut the inverted third line q q in J and the third line y Y in l, 
then L b' must be a support tangent. Produce l b' to meet 
the first third line of the span b c in l', then j l' is the left- 
hand mid link; and then join fb' and produce it to meet 
J l' in f', then f' will be a fixed point on the mid links of the 
second span. This is shown as follows — 

Let the vertical through f' be at distances z-^, 23 from the 
third lines. 

Then the triangles f' j n, f' k' l' are similar. 

• • k'l' 22 ^^ 



and triangles b' k' l', b' k l are similar. 

(2) 



k' 1/ I2 



further, the triangles f l k, f j n are similar. 



- , (3) 

JN /2 



FIXED AND CONTINUOUS BEAMS 455 



Multiplying together ( 


1), (2) and (3), we get 


1 




X r 




^2 12 


h 


. ^1 ~ ih 


_ h /2 




^2 


hh 




also Z^ -T ^2 


h + h 

3 




.-. Zi - 1Z2 


= ~Z2 + ^h 




. -^2 + ih 


_^l/2 




^2 


~hfl 






= 1 + ^^^2 





•' ' ^2 — 1~T~^JIjT ~ constant. 
h / 2 + ^2 /I 

. • . r' is a fixed point. 

In this way a number of fixed points right along the various 
spans can be found as hereinafter further explained. 

A fixed point is found at the terminal spans, as follows — 

Case 1. Freely Supported End. — The end B.M. here 
must be zero, therefore support tangent and mid link must be 
collinear, so that a' is the first fixed point. 

Case 2. Built-in or Fixed End. — Support tangent is 
horizontal, so that first fixed point is where horizontal through 
a' cuts the first third line. 

Graphical Construction for any Given Case. — We 
are now in a position to set out the construction for obtaining 
the B.M. diagram, which is as follows — 

Draw the free B.M. diagrams and the third lines, the 
inverted third lines and the centroid verticals. Fig. 215 
shows a continuous beam of three spans, one end being freely 
supported and the other fixed, x x representing the left-hand 
third lines, Y Y the right-hand third lines, q q the inverted 
third lines, G G the centroid verticals. 

Now draw the cross lines at the bottom of the paper, such 
lines being obtained by setting down the areas S^, Sg, etc., 
of the free B.M. curves on vertical lines at each side of the 
centroid verticals at distances representing the value of E I 



456 



THE STRENGTH OF MATERIALS 




CO 

C 

a, 






eS 

« 

CO 

o 

•S 

'43 

o 
O 



eg 

to 



e3 
o 

a, 
O 



6 



^^^ 



FIXED AND CONTINUOUS BEAMS 457 

reduced in some convenient ratio, the scale of E I being the 
same as that of the areas. If the support moments only are 
required and not the deflections, and E I is the same for each 
span, E I need not be calculated, any convenient polar distance 
being taken. 

P, P^, and Pg are the intersections of the cross lines. 

Now find the fixed points. The end a is fixed, so that f is 
the first fixed point ; now set down r r' equal to the intercept 
■ / /i on the cross lines and draw any line r' J^ to the inverted 
third line, cutting y Y in l ; join L b' and produce to meet the 
third line x^ x^ in l^ ; then the intersection of l^ j^ and e' b' 
gives the fixed point Fj^ on the second span. This is repeated 
as shown, and the points f^', Fg, Fg' found. 

^ow start the other end d. This is freely supported, 
therefore, as we have seen before, d' is the first fixed point Hg. 
By means of the cross lines, we then get the corresponding 
fixed point Hg', and by repeating the same construction as 
for the points F, we get a number of other fixed points, 
Hj', H^, h', h. The mid links and support tangents are now 
drawn in, and there will be two checks on the accuracy of 
the construction, viz. — 

(a) Mid links must meet on centroid verticals. 

[b) When adjacent mid links are joined, they must pass 

through points of support. 
Now, from the points 1, 2, etc., on the cross lines, draw 
parallels to the mid links, and obtain the poles r, r^, Rg and 
then draw parallels to the support tangents, thus obtaining 
the points 0, 3, etc. Then 

and so on, the support moments then being set up and the 
true B.M. curve for the continuous beam thus being found. 

Another interesting graphical method of finding the support 
moments in a continuous beam has been devised by Professor 
Claxton Eidler, and will be found in his book on Bridge 
Construction. 



458 THE STRENGTH OF MATERIALS 

Advantages and Disadvantages of Continuous 
Beams. — It will be seen by considering the B.M. diagrams 
for continuous beams that the maximum B.M. is less than 
that which would occur if a number of separate simply 
supported beams were placed across the same supports (ex- 
cept in the case of two uniformly loaded equal spans, when 
it is the same), and that such maximum B.M. occurs at the 
abutments. The principal disadvantages are — 

(a) It is not easy to ensure all the supports remaining at 

exactly the same level. 
(6) The method of calculation of the stresses assumes that 
the beam is of uniform cross section throughout, this 
condition not being an economical one. Experimental 
investigations in Germany have shown that if the 
beam is not of uniform cross section, the method 
described may still be employed without great error, 
(c) In the case of rolling loads, which occur frequently in 
bridge design, the • calculations are much more 
difficult than in the case of separate spans. 

Beams Fixed at one End and Freely Supported at 
the Other. — If a beam is fixed at one end and freely sup- 
ported at the other, the B.M. and shear diagrams will be the 
same as for the half of a continuous beam of two equal spans 
of the same span as the given beam, and loaded in the same 
manner. 

This is because fixing the end of a beam makes such end 
horizontal, and this is what happens at the central support 
of a continuous beam with two equal spans loaded in the 
same manner. The consideration of the following two 
standard cases should make this clear. 

(a) Beam Fixed at one End and Freely Supported at 
THE Other, Subjected to a Uniform Load. — The 
B.M. and shear diagrams in this case are the same 
as for one span of the first case of continuous beams 
that we have considered, and will therefore be as 
shown in Fig. 216. 



FIXED AND CONTINUOUS BEAMS 



459 



(6) Beam Fixed at one End and Freely Supported at 

THE Other, Subjected to a Central Load. — Let 

the central load be W and the span I. 

Then, if b is the fixed end, a the freely supported end, and 

a' the imaginary freely supported end existing beyond the 

fixed end, we have, by the Theorem of Three Moments, 



MJ + 2M,(? + Z) + M,Z = 6|^ix|x2^^ + ^x|x^^ 
Now M, = M,, = 

.-. 2M,..2? = 6 



W Z3 W ^3^ 



. M. 



IQl 
16 



16 Z J 



} 





Fi€t. 216. Fig. 217. 

Beams Fixed at one End and Supported at the Other, 



The B.M. diagram then comes as shown in Fig. 217. 

To get the shear diagram we first work out the reactions. 

j^ _ W , M, - M„ 



. ' . R,. 



~ 2^ 


I 


W 

2 


3WZ 

IQl 


5W 




16 




11 W 





16 



The shear diagram then comes as shown in the figure. 



460 



THE STRENGTH OF :\L\TERIALS 



(c) to find the maximum deflection for a uniformly 
Loaded Beam fixed at one End and Supported 
AT the Other. 

The bending moment diagram for this case is as shown 

shaded in Fig. 218. The curve b d c is a parabola of height -^ 

where iv is the load per unit length of the beam, this being the 
B.M. diagram for the downward miiform load on the canti- 




FiG. 218. — Deflections of Beams. 



lever, and a j is equal to 



3 ic P 



! J B being a straight line ; this 

3 ic I 
being the B.M. diagram for the reaction at b, which is ^— 

8 

Our first problem is to find the point n at which the deflec- 
tion has its maximum value. Consider the position Aj^n of 
the imaginar}^ cable. The forces acting on it are a horizontal 
tension equal to e i at n and an equal horizontal tension at 
the point a^, since the beam must be horizontal at the fixed 
end a; also an upward vertical force equal to the negative 



FIXED AND CONTINUOUS BEAMS 461 

area cjb, and a downward vertical force equal to the positive 
area d f g. 

If these forces are in equilibrium, since the horizontal forces 
are equal and opposite, the vertical forces are also equal and 
opposite, so that we get the following rule — 

The maximum deflection will occur at the point where the area 
D F G ^5 equal to the area d J c. 

This is the same as saying that the area a h g J is equal to 
the area a h f c. 

Now, if H B = X and a b = Z 

HG_AJ ^ _a;AJ 

— y— . * . H G y- 

X I I 

Area a h g J = -^ (a J + g h) 

V '- X I -I , X 

A J 1 + 



.2 ^" V Z 
{I - x) {I + x ) SwP 
'2~~ I ' 8 

3wl 



16 



(P -x^) (1) 



Also Area a h f c == Area a b d c — Area f h b 

1 1 

= ;^AC.AB — „ FH.HB 

Iwl^ 1 W X^ 
^ 3~2 "" 3 ~T~ 

If (1) = (2) 



(2) 



^^ (l^ - ^') =^ I (^' - ^') • 

Factorising, we get 

^^Q^ {I + x)(l- X) =^~{l- ^) (l' + l^ + ^^') 

w 
. ■ . dividing through by ^ (^ — ^) ^^^ multiplying across 

we get 

9?2 + 9?a: = 8Z2 + 8Za: + 8x2 

i.e. Sx^ -Ix -l^ = (3) 



462 THE STRENGTH OF MATERIALS 

The general solution of this quadratic equation is 



X ^ 

_ I (1 ± V33) 
16 

The negative value is inadmissible 

.•.. = L(L+/33)^. ^2 nearly. 

. • . The maximum deflection occurs at a distance = '422 1 from 
the simply supported end. 

We now proceed to find the maximum deflection S by con- 
sidering the stability of the portion n Bj of the imaginary 
cable. The forces acting on it are a tension at B, the horizontal 
tension E I at n, and the area of the bending moment diagram 

B F G. 

By taking moments about the point b^, we eliminate the 
tension at this point and get E I x 8 = moment about B^ of 
area b F G. 

Now, this area is made up of the difference between the 
A B H G and the parabola b h f. 

X 

T 

__ I x^ S wl^ _ 3w x^l 



The area of the A = jGH.BH = J.^.Aj.a; 



M • 8 16 

2 X 
The centroid of the A is at distance -^ from b 

, p ^ , , Zwx^l2x w x^ I 

.' . moment oi A about b, — — ^tt^ — . -^ = --;: — 

lb 3 8 

The area of the parabola = J f h . b H 

_ I w x^ _ w x^ 

The centroid of the jDarabola is at distance — ~ from b. 

zv X 3 X 
. • . moment of parabola about B, = —7, . -^ 

o 4 

w x^ 



FIXED AND CONTINUOUS BEAMS 463 

.* . moment about b^ of area b f g 

w x^l w x^ 



E I X 8 = -j^ (Z - a;) 

putting X = '422 I 

^ T ^ w X -4223 P (-578 I) 
. • . EI X S = ^ — ^^ ^ 

o 

= -00543 w Z* 
, -00543 w l^ 

•*• ^^ — Eir~ 

putting wl = total load = W 

. -00543 W P 

^ = EI 

_ WP 
184 EI 

For a uniformly loaded beam, simply supported at each 

5 W Z^ 
end, we should get 8 = „- . t^ j , while for one similarly loaded, 

WP 

but fixed at each end- we should get 8 = ttttttTtj so that 

^ 384 E I 

we see that, in the case under consideration the deflection is 

between these two values. This is, of course, what one would 

expect. 

The same method may be applied to the case of an isolated 

central load W on a beam similar fixed. In this case the 

maximum deflection = >_ and occurs at ^7^ from the 

48 VS E I a/S 

simply supported end. 

We will conclude this chapter with a further number of 
worked examples of fixed and continuous beams. 

Worked Examples. — (1) A beam of 20 ft. span is built-in 
at one end and is supported at a point 5 feet from the other end. 
Draw the B.M. and shear diagrams for a uniform load of J ton 
per foot run. 



464 



THE STRENGTH OF MATERIALS 



Let AB (Fig. 219) be the beam, fixed at the end a and 
supported at the point c. 

The portion b c of the beam acts as a cantilever, and there- 



fore the B.M. at c = M, 



1 5x5 

^ X ^ = 625 ft. tons. 



To find the B.M. at a, we imagine a span a c' exactly similar 
to A c to exist within the wall. 




Fig. 219. 

Then, by the Theorem of Three Moments, we have 
M, X 15 4- 2 M, (15 + 15) + M, . 15 = ^ (15^ + 15^) 

o 

But M,. = M, - 6-25 

.-. 60 M, + 30 X 6-25 - ^(2 x 15^) 

o 

.-. 4M, + 12-5 = ^f 
4 

1 ^12 

... 4M, - , - 12 5 
4 

= 56-25 - 12-5 =43-75 

.■ . M, = 10-94 ft. tons nearly. 

The B.M. diagram is then as shown in the figure. To get 
the reaction at c we proceed exactly as in the case of con- 
tinuous beams 



FIXED AND CONTINUOUS BEAMS 465 

. ^ 1 15 , M, - M^ 1 5 M, -M, 

I.e. J:^, - 2 . 2 "^ '""'15 '^2'2 "^ 5 

= 3-75 - -31 + 1-25 + 1-25 
= 3-44 + 2-5 
= 5*94 tons. 

The shear diagram then comes as shown in the figure. 

(2) A roiled joist is firmly built-in at one end, and the other 
end rests freely on the top of a cast-iron column. The span of 
the joist is 16 feet, and it carries a single load of 10 tons, 12 feet 




Fig. 220. — Example of Beam Fixed at one End and Supported 

at Other. 

from the column ends. Determine the reaction on the column, 
and draw the B.M. and shear diagrams. {B.Sc. Lond.) 

Let A B represent the beam, fixed at the end a, the load 
being at the point c (Fig. 220). 

Then the free B.M. diagram is a triangle a d b, c d being 
equal to 

30 ft. tons. 



Wa& 10 X 12 X 4 



I 16 

Then area of B.M. diagram 

= J X 30 X 16 - 240 sq. ft. tons. 
The centroid g of the B.M. diagram occurs at a distance 

HH 



466 . THE STRENGTH OF MATERLILS 

J E c from E the centre of the beam, i. e. at a distance 9J ft. 
from A. 

Then, imagming a span exactly similar to ab to exist 
beyond the fixed end, we have, by the Theorem of Three 
Moments 

16 M3 + 2 M. (16 + 16) + 16 M, = 6 {^^\l^'' + ^^'} 

M3, = M, = 

.. T^ 6 X 2 X 240 X 28 ^ ^,^ 

64 M, = — ^ = 7 X 240 

16 X 3 

M = 7 X 240 _ 210 
64 " 8 
= 26-25 ft. tons. 

The reaction on the end b for a freely supported beam 

= ^B = — T^ — = 2*5 tons. 
Id 

. '. In this case R^ = r^ H -~~ 



= 2-5 + 



I 
- 26-25 



16 
= 2-5 - 1-64 

= *86 tons. 

(3) A continuous girder consists of two unequal spans of 
100 ft. and 120 ft. respectively. The girder is 300 ft. long and 
overhangs the end supports at each end, and is loaded as shown 
{Fig. 221). Draw the B.M. and shear diagrams and show the 
points of inflexion and magnitude of the supporting forces. 
{B.Sc. Lond.) 

In this case the end pieces A B, d e act as cantilevers. 

,, 40 X IJ X 40 , ^^^ J., . 
.• . M„ = ^ = 1,200 ft. tons. 

40x2x40 ^1^600 ft. tons. 



. FIXED AND CONTINUOUS BEAMS 



467 



The free B.M. curve for span b c is a parabola with maximun' 

,. ^ li X 100 X 100 , ^_^ „^ ^ 
ordinate = -^ ~ = 1,875 ft. tons. 

The free B.M. curve for span c d is a parabola with maxi- 

,. , 2 X 120 X 120 o^AA^^ ^ 
mum ordmate = = 3,600 ft. tons. 



lit^ns l^cr Ft. run ^*"^ M ^'trun 

nnmoononnoOnnOOOOOOO 



^ 



40-»- 



lOC 




B.M. Dioqram 



12.0 



Shear l^loaram 
Fig. 221. 



Da 

— >■ 



■<- 40 -> 




Then applying the Theorem of Three Moments we have 

100 M„ + 2 M, (100 + 120) + 120 M, = i (1 J X 100^ + 2 x 120^) 

.-. 120,000 + 440 Mo + 192,000 = 375,000 + 864,000 

440 M, = 927,000 

M, = 2,107 ft. tons nearly. 



468 



THE STRENGTH OF MATERIALS 



We now proceed to the determination of the reactions, 
o - 1 V <in V 1 1 M,.-M, 1 1 M„-M, 



= 30 + 30 + 75 - 9-07 
- 60 + 65-93 



= 125-93 tons 

M, - M„ 



Re = ^ X 100 X 1^ + ^^^^J" + ; X 2 X 120 + ^2^ 

= 75 + 9-07 + 120 + 4-22 

= 84-07 + 124-22 = 208-29 „ 

R., = i X 2 X 120 + ^'i7/'^ + ^ X 2 X 40 + ^" ~^^ 

= 120 - 4-22 + 40 + 40 

= 115-78 + 80 = 195-78 „ 

Total .. 530 tons 




The shear diagrams then come as shown on the figure, and 
the points g h k l are the points of inflection. 

(4) A continuous beam of total length L has three spans and is 
uniformly loaded. Find the most economical arrangement of the 
spans. 

It follows from symmetry that in the best arrangement the 
two end spans will be equal. Let the epd spans be of length /j 
and the centre span of length I2 (Fig. 222). 

Then L = 1^ + 21^ 

Now by the Theorem of Three Moments 



w 



M Ji + 2 M,. {I, + I,) + M J2 = : (^1' + ^2') 



FIXED AND CONTINUOUS BEAMS 469 

From symmetf y M,. = M„ 
also M^ = 

We now require to find the relation between l^ and I2 to 
make M„ a minimum, and then see if M,, is greater than 
the intermediate B.M.s : if so, this relation will give us the 
most economical arrangement. 

M = ^ 
" (2 ^1 + 3 l^) 

Now ?2 = L - 2 ?i 

(3 L - 6 ?i + 2 Zi) 

This will be a maximum when , ,-" = 

a l-^ 

i. e. when (3 L - 4 ^i) (- 21 l^^ + 24 /^ L - 6 L^) 

+ 4 (L3 - 6 L2 ?i + 12 Ij^L - 1 l^^) == 
i. e. 56 l^^ - 111 L Zi2 + 72 ^^ L^ - 14 L^ = 

The solution of this equation will be found to be l^ = -35 L, 
such solution being found by plotting. 

Thus we see that the least value of the support moments 
occurs when the end spans are each "35 L and the centre -3 L. 
In this case the intermediate B.M.s are less than the support 
moments, so that this gives the most economical arrangement. 



CHAPTER XVI 

* DISTRIBUTION OF SHEAR STRESSES IN BEAMS 

When a beam is deflected there is a horizontal * shearing 
stress at every point of the beam, resisting the shding of one 
layer over the other. We have already shown (p. 10) that in 
an elastic material a shear stress must always be accompanied 
by a shear stress of equal intensity at right angles to it ; in 
the case of the beam we see that the horizontal and vertical 
shearing stresses at any point of a beam are equal. Now the 
total shearing force over any vertical cross section of a beam 
must be equal to the shearing force, obtained, as in previous 
chapters, by considering the forces on the beam; but the 
intensity of stress will not be the same across the section, so 
that by dividing the shearing force S by the area of the cross 
section A, as is commonly done, we do not get the maximum 
shear stress. 

The existence of the horizontal shearing stress can be seen 
clearly from the following diagrammatic representation. 
Fig. 223 (a) shows a short beam deflected under some loading. 
Now imagine the beam to be replaced by a number of plates 
placed one above the other. They then take the form shown 
at (b) on the figure, the plates sliding one over the other as 
shown. The second case will not be nearly as strong as the 
first case, and it is clear that in case (a) there must be stresses 
tending to make one layer slide over the other. 

* We will assume through this investigation that the beam is 
horizontal. If it is not, the words "parallel to the axis of the beam" 
and " perpendicular to the axis of the beam" should be substituted 
for "horizontal" and "vertical." 

470 



DISTRIBUTION OF SHEAR STRESSES 471 

We will now obtain an expression for finding the shearing 
stress at any point of a beam, and will consider later certain 
special cases. 




® 




® 



Fig. 223. — Horizontar Shear in Beams. 




Fig. 224. — Distribution of Shear. 

General Case. — Let ab, a^Bi (Fig. 224), be two cross 
sections of a beam at a short distance x apart, and let the 
cross section of such beam be symmetrical about a vertical 
axis, and let the loading be wholly transverse. Then E c g and 
EiCiGi, as we have previously seen, give the intensities of 



472 THE STRENGTH OF IVIATERIALS 

transverse stress at any point. Now consider the portion of 
the section ab above any line d d. Consider an element of 
area a at a point p at distance p x from the neutral axis. 
Then we have by the theory of bending that the intensity 

of stress at p = /^ = j , where M is the B.M. at the 

point and I the second moment of the section. 

.• . Force on element a = /^ x a = " ^ . a 

. ■ . Total force on area above d d = 2 j . a 

M 

= ^ X first moment of area above d d about N.A. 

= J X a.y (1) 

Where a is the area above d d and y the distance of its centroid 
from the N.A. 

Similarly taking the section a^ b^ and taking the force above 

a line Di Di we have 

M 

Total force on area above d^ d^ = Fj^ = -p - x a^ y^ 

Now, if X is small, and the beam has no abrupt change in 
cross section, we may put a = a^, y = y-^, and I = Ii 

. ...F-F, = <^^i>^ (2) 

Now this difference in transverse force is the shearing force 
which has to be carried along the line d d^. We will write this 

^ _^ ^ (M - Ml) a^y^x 
^ X I 

M — M 
Then, if x is very small ^ is the rate of increase or 

decrease of the B.M., and this we have shown to be equal to 
the shearing force S at the given point. 

.-. We have ^ - ^^ = ^^ ""^y^^ (3) 



DISTRIBUTION OF SHEAR STRESSES 473 
Now the area over which this shearing force acts is equal to 

B^D X T) B = -D T> X X = X X b. 

. ' . Mean shearing stress along d d 



D 


= 


F 


b X 






= 


S 


X a.y . 


. x 






bx .1 




Sj, 


= 


S 


.a.y 
1.6 • 





(4) 

We can express this in terms of the mean stress m = -^ over 

the whole section as follows — 

_S . a .y 
^" ~ AVkH 

a.y ,_, 

We may call j~ the shear coefficient. 

It will be noted that a x y increases up to the neutral axis 
and then decreases, because the first moment of the area below 
the N.A. is negative. 

We thus see that the shear stress is a maximum at the neutral 
axis. 

It must be remembered that 5^ gives only the mean shear 
stress along d d. This stress is not uniform along d d, but 
for sections which are narrow at the neutral axis, the sections 
used in practice generally falling under this head, the maximum 
shear along the neutral axis will be not much greater than 
the value of s^ at the neutral axis as given by the above 
result. For sections like the square and the circle the maxi- 
mum shear along d d will be from 5-10 % greater than the 
mean shear, while for sections such as an oblate ellipse or a 
broad rectangle the difference may amount to as much as 
25 %. It is beyond our present scope to go further into the 
question as to the variation of shear stress along d d, but 
we should remember that such stress is not uniform; the 
maximum stress for various cases has been worked out by 
St. Venant. 



474 



THE STRENGTH OF MATERIALS 



Consider the following special cases (Figs. 225, 226). 
(1) Rectangular Section. — Mean shear along a line at 
distance x from N.A. of a rectangle of height h and breadth b 



_ _ ^ -y 

- -5. - w . -p-^- 



In this case a 



-x]b 



y = X + 



l/h 



k^ = 



12 



2V2 



X = 



+ x 




fxeclarkjle 



Parabola 




Cinclt 



Fig. 225. 



s. 



m .[^ — xjh . ^[^^^ X 



12 

6 m ( . — x^ 



= 6 m ( J - ^, 



_ 3m /, _4:X^ 

~ 2 \ 'W. 

This depends on x^, so that the curve showing the mean 

shear stress at various depths will be a parabola. The 

maximum value of s,. occurs when a: = 0, ^. e. at the neutral 



DISTRIBUTION OF SHEAR STRESSES 475 

axis. This gives 5<, = —^ — 1*5 m. Thus we see that in a 

rectangular beam the maximum shear stress occurs at the 

centre, and is equal to 1*5 times the shearing force divided 

by the area of the section. 

(2) Circular Section. — This case is not quite so simple as 

the previous case, but we can find the shear stress at the N.A. 

simply as follows — 

In this case we have 

7rD2 

2D 

6 =D 

ttD^ 2D 

8 • Stt 

. • . 5„ . = m 



D2 
16 
4 w 



.D 
= 1-33 m 



So that the mean shear stress along the N.A. is 1 J times the 
mean shear stress over the whole section. 

In this case it is interesting to note that the maximum shear 
stress along the N.A. is 1*45 m. 

(3) Pipe Section. — Let a thin pipe be of mean diameter 
D and thickness t (Fig. 226). 

Then a = —^ 

D 

y ^^ 

IT 

D2 

^ ~ 8 

6=2^ 

ttD^ D 

. • . 5v A = 'W* X -:^ = 2 m 

^ X 2* 



476 



THE STRENGTH OF MATERIALS 



So that the mean stress shear along the N.A. is twice the 
mean shear stress over the whole section. 

(4) I Section. — To calculate the proportion of the shearing 
force carried by the flanges and web, respectively. 

Take a beam of I section of breadth h and height h, and 
let the thickness of the flanges and the web be t and w, 
respectively. 

First consider a horizontal line p p in the flange at distance 
X from the top edge (Fig. 226). 




i^ 



Pi 



P-h7--f — ?4^ 






uy 



T 



-?, 



B 



\K 



Ml 



Fig. 226. 



Then mean shear along p p = m . ^^g ; 



s,. = 



m .b X {h — x) 
" 6 F 2 



— 9 y^ 2 V ^ ^ ) 



(1) 



This depends on x^, so that the curve showing the variation 
of stress is a parabola. 

When X = t, i. e. at the junction of web and flange, 



St = 2^2 (^ ^ - ^') 



(2) 



Now consider a horizontal line Pj^ p^ in the web at distance 
x^ from the top. 

Then mean shear along pj p^ — -' '— 



DISTRIBUTION OF SHEAR STRESSES 477 



In this case 

ay = first moment of area above p^ p^ about N.A 

_b t {h — t) ^ ^ ,_ ^^ \h /^ ^ , ^1 — ^ 



+ w{x,-t)\^^-[t+^-^) 



_b t {h — t) w {Xi — t) (h — x^ — t) 

- 2 + 2~' 

also b = w in general expression for shear stress. 



'n 9 1,2 



m fbt{h — t) (% — t) (h — x^ — t) 



2 k^ I w w 

_ m , mt (h — t) {b — w) .ox 

- 2 p (^ ^1 - ^-1 ) + -^2k^w « ^^^ 

The second term of this expression is constant for all values 
of x-^^ and the first term is the shear stress which would occur 
if the fianges extended down to p^ p^. 

We thus see that the diagram of distribution stress is 
obtained as follows — 

First draw a parabola a k d, the centre ordinate J K of 

h 
which is obtained by putting a; = ^ in equation (1). 

I.e. JK = 



2PV2 47 8 F 

At the points b and c corresponding to the inside edges 
of the flanges set out g e and h f equal to the expression 

o r.2 ^^^ re -draw the portion g k h of the 

parabola between the points e and f, then the curve 
A G E L F H D givcs the shear stress at the various depths 
of the cross section. 

Then total shear carried by web is equal to area of piece 
B E L F c of curve multiplied by width of web. 

Now take the case in which ^ = v^ and w — ^ and b =^ ^, 



478 THE STRENGTH OF MATERIALS 

this being about the proportions for a rolled steel joist, then 

BG =s,= ^^^^ {h t - f") 

_ m /h^ h^ \ 
^ 2k^ VlO ~ 100/ 
_ m 9 ^^ 
"" 2 A:2 • 100 

_ m /h^ 9 h^\ __ m 16 h^ _ m 4 A^ 
•" • ^ ^ "" 2 F [J - 100 j ~ 2k^ • loo ~ 2l2 • "25" 

also mt{h-t){h -w) 

also GE - ^^^~ 

_ m h 9h 9h 20 
~ 2 F • rd • 10 * 2¥ ^ 7j 
_ m Slh^ 
~ 2T2 • 100' 
_ m 9 h^ m 81 h^ 
''' ^^^ 2k^' 100 '^ 2 F • Too 
m 9h^ 



2 k^ ' 10 
. • . Area of curve belfc =BcfBE +^mk 

_ 4^ m /9h^ , §^^ (A) 

" 5 • 2 P VTO ^ 75 y • • • • ^ ^ 

Now m this case i = -j^ yo 

= ^ _ ?i* (thy L 

~ 24 20 • V 5 j • 12 
= -0417 h* - -0192 h^ 
= 0225 h^ 
The area of the section = b h — {b — w) {h — 2 t) 
__ 7^2 _ 9^ 4^ 

~ 2 20 • 5 

= •14^2 

r. k^= ^=-?l^f = '1608^' 
A T4/j2 

Returning to equation (4) we get area of curve belfc 

4 m A r9 y^2 8 ^2 

+ 



10 F I 10 ' 75 



DISTRIBUTION OF SHEAR STRESSES 479 

" 10 X -leosl^ ' 

, 4 X 1007 
^^^•^^-1-608 
= 2-505 m^ (5) 

. • . Shear carried by web = 2*505 mil x width of web 

= 2-505 mh X ^^ 

= -1262 mh^ (6) 

Now area of whole section = • 14 ^^ 

.'. Total shear S on section = "W/i^ x m 

Shear carried by web _ • 1252 _ r.^ ^ o/ 
•*• Tot^rshear O^ ~ /°* 

It is commonly assumed in practice that in plate and box 
girders the whole of the shear is carried by the web, and the 
above calculation shows that in an I beam, in which the 
flanges are larger in proportion to the depth than in most 
plate and box girders, this is true within 10 %. 

It must, however, be remembered that in girders built up 
of joists and plates, such as the comparatively shallow and 
heavy girders used in buildings, this assumption will not be 
so nearly true. 

Shear Stresses in Reinforced Concrete Beams. — ^The 
usual treatment is as follows — 

Consider two vertical sections of a reinforced concrete beam 
made at points ab a short distance x apart (Fig. 227), the 
section not changing appreciably from a to b. At the point 
A, the total stresses due to the bending moment are C and T, 
and at b they are C and T' ; then if the corresponding bending 
moments are B and B', we see that 

T = C = ? 

a 

a 
...T-T'=^^ ~ (1) 

Adhesion Stress due to Shear. — But T — T' is the 



480 



THE STRENGTH OF J^IATERIALS 



difference in the pulls in the reinforcement at the two points ; 
that is, it is the force which tends to pull the reinforcement 
out of the concrete. 



T -^r 

X 



adhesive or shear force per unit length 



X . a 



B 



T 



a> 



X. 



T' 



^JS 



a. 



Fig. 22: 



but 



B -B' 



X 



the rate of change of the bending moment 



shearing force = S 

the adhesive stress per sq. in. and 

the total perimeter of the reinforcement, 



and if /„ 
O 
we have 

/„ X O = adhesive stress per unit length = 



T - T' 



X 



■■ /„ = 



s 

O .a 



(2) 



DISTRIBUTION OF SHEAR STRESSES 481 

In the case of rectangular beams with tension reinforce- 
ment only or with double reinforcement where the top re- 
inforcement is placed at J depth of N.A. a = (d — ^ 

. our formula becomes /„ = — ; -^ (3) 



ou - 



This deals Avith what is known as the horizontal shear as 
regards the adhesion between steel and tension. 

Shear Stress in Concrete.- — In addition to this we have 







/ 


> 




^=!*S^ 




/ 


^ 




V 




\ 




1 

/ 


/ 


\ 




n 


\ 




\ 




\ 






1 


\ 




d. 


\ 




\ 






\ 




* 










/ 




/ 


CL 


< ^ > 


/ 


\ 




ccb 


/ 




/ 




/ 




/ 


\ \ 




i/y//////} v//f/ //////// 

/ 













Fig. 228. 



to consider the shear stress in the concrete itself, w^hich will 
be constant from the reinforcement to the N.A. and will 
then vary towards the top as indicated in Fig. 228. 

The difference of pulls T — T' is distributed over a hori- 
zontal rectangular area one side of which is x and the other 
side of which is b, the breadth of the beam. 

. • . If 5 is the shear stress we shall have 

s xb X X = T ~-T 
but T - T' == ^'^ (from (1)) 
(B - B') 



s = 



X . a .b 



(4) 



II 



482 



THE STRENGTH OF IVIATERIALS 



But as before we may put — - — = shearing force S 



X 



S 

a 
and for reactangular beams as above 

S 



,9 = 



b d 



n 



(5) 
(6) 



GRAPHICAL TREATMENT FOR FINDING DISTRIBUTION 
OF SHEAR STRESS ON A CROSS SECTION 

Consider the section, composed of joists and plates, shown 
in Fig. 229. The first step is to " mass the section up " about 




p \Q ft ,- — ; 


1^ 


'^L^ .-' ir 


i 




_j' 


/I' 


'■1 

1 


A 




IS B 


\ 






-^ c- 


1 
1 
1 




JC 


«- ^ 


1 
1 

1 


^ ^ 





r 



Fig. 229. 

a vertical centre line : this is done by drawing horizontal lines 
across the joists, and adding on each side of the centre joist the 
corresponding horizontal ordinate of the outside joists. This 
gives the section shown in the figure {i.e.a d = ah -]-hc -{- c d). 
Consider any line p p. We have shown that the mean shear 



stress along p p = <s, 



m 



Ic^h' 



Now a . y = first moment of area above p p about neutral 
axis X X. Draw the first moment curve of the section above x x 
about the line x x, as explained on p. 176. As the section is 
symmetrical about a vertical axis, we need draw the curve 
for one half of the area only, x q c being such curve. 



DISTRIBUTION OF SHEAR STRESSES 483 

Then a x y = 2 area J x Q n x h, 

.*. Mean shear along p Pi = ^„- . • = 

Now find the sum-curve j r s of the first moment curve, 
taking the polar distance p = , . 

Then n e. x ^^ = area of first moment curve above p p 

N R X A^2 



'h 



= area j x q n 



.*. Mean shear along p p = ^-p . ~j. — X h 

2 N R 

= m . — J — 



But 6=pp = 2np 

. • . Mean shear along p p = m . 

^ NP 

Then the maximum shear stress, which occurs at the neutral 

. . c s. 

axis, IS m . 

c B 

Note. — Fig. 229 is diagrammatic only and is not drawn to 
scale. The student should work this case as an example, 
taking the plates 20'' x i'' and W x 6'' beams. For 
accuracy the drawing should be done to a large scale. 

Deflection of a Beam due to Shear. — In considering 
the deflections of beams up to the present we have dealt 
only with the deflection due to the bending moment. We 
will now see to what extent the deflection due to shear is 
comparable with that due to the bending moment. 

Let c c (Fig. 230) represent a short length x of the centre 
line of a beam subjected to a shearing stress s. 

Then the shear causes the line c c to take the position c c^, 
the slope being a. 

Then if G is the shear modulus, we have o- = ^ 

The deflection c c^ of the short length of beam is equal to 
X X cr, as o- is small. 



484 



THE STRENGTH OF MATERIALS 



Deflection of short length x of beam = 
. • . Total deflection due to shear = ]S 



X X s 

X X s 



G 



a 11 
Now we have shown that s = m . j-^^ where m 



S 



, S being 



he shearing force at the point, and A the area of the section. 

ft fj 
If the section is uniform along its length, r-^^ will be 

constant and equal to, say, /?. 



s 


i 










c 




C 




cr 


C-. 


""" — ~. , 






y 


s 



M 




M, 


P 




P, 


y 


tm T^ t- 


r; 


— .^^ r 




c 




<^/ 



Fig. 230. 
. • . We have : deflection due to shear 



Fig. 231. 

/5 .S 



2 X , 
AG 



A.G 

%x.S 



But 2 . a; S = area of shear curve up to given point 

= B.M. at point 
= M 

. • . Deflection due to shear = /x = -^^ . M 

Now consider the following special cases — 
(1) Isolated Central Load. 

(3 Wl 
Deflection at centre = />t = -^-p . — ^ 



(i: 



DISTRIBUTION OF SHEAR STRESSES 485 
As we have previously shown, the deflection 8 in this case 

due to B.M. is equal to .^ ^-i x 
^ 48 E I 

•*• a ~ AG • 4 • 48 EI 

_ 12E ^I 

~ G • A Z2 

E 5 

Taking ^ = ^ and noting that I = A A;^ 

1=30^^1-' (2) 

(2) Continuous Loading. 

In this case //, = ^'-^^ • o 
5 W Z3 



384 E I 

/x _48j8 E I 
''' S~ 5 • G • A> 

Taking ^ = o as before, 

^ = 24/3.^' (3) 

For rectangular section jS = 1'5 and k^ = y^, h being the 
depth of the beam. 

.-. (2) becomes^ = 3-75 (y)' 



(3) becomes^ ^ ^ ( 



I 



It follows from this that if y = ^^, the deflection due to 

shear is 3* 75 per cent, and 3 per cent, respectively of that due 
to B.M. in the two cases. 

We see, therefore, that for solid rectangular beams in which 
the span is more than 10 times the depth, the deflection due 
to shear is negligible. 

It must, however, be remembered that for rolled joists, 



486 THE STRENGTH OF IVIATERTALS 

plate girders, and the like, the deflection due to shear will be 
quite appreciable for sections which are deep compared with 
their span. Bridge engineers often state that the deflection 
of a bridge is more than the calculated deflection. Part of 
this difference may be due to the giving in the riveted con- 
nections, but certainly the measured deflection would agree 
better with the calculated deflection if the latter included the 
shear deflection. It has been suggested that this could be 
remedied by taking E about 10,000 tons per sq. in. instead 
of 12,500 in the ordinary deflection formula. 

It should also be noted that we have taken only the strain 
due to the maximum shear stress, neglecting the fact that it 
is variable. This gives results a little too high, but is better 
than taking the mean shear stress. 

Distortion of Gross Section of Beam due to Shear, 
etc. — In finding an expression for the relation between the 
stresses and the B.M. on a beam, we made use of Bernoulli's 
assumption that the cross section remains plane after bending. 

The two causes tending to distort the cross section are 
(1) shear stress, (2) differences in lateral compression due to 
extension in fibres. 

Consider two cross sections of a beam at distance x (Fig. 231) 
apart, and let the B.M. at the sections be M and M^ respec- 
tively, and consider points p and p^ at distance y from the 
centre line, the section being the same at the two points. 

Then stress at p = -y" , at p^ = — -i — 

T ^ 1 . ^ . ^ M?/ ^ Ml?/ 

.' . Lrateral compression strani at p = 77 -nrj^ ^'^ ^i =^ "^ -c« j 

stress 
because longitudinal strain, = ' — :^ — and lateral or transverse 

strain = 7; x longitudinal strain. 

. • . Difference in lateral compression strain = ^j . (M^ — M) y 

.'.On a short length d y of the section, the difference in 

lateral compression = p' p^" = J/j . (M^ — M.).y.dy 



DISTRIBUTION OF SHEAR STRESSES 487 

.• . a = slope of p Pi = ^^ = -g'j • --'— ^ • 2/ • ^ 2/ 
but we have shown that when x is ven^ small 
^—^ = the shearing force S 



X 



j7 J .^ .y..dy 



To find the total change in angle between any section and 
the line originally parallel to the centre line, we must add 
all the elementary changes in angle. 

S 



Total change = 


: 


= Eiy^ 


.dy 








2EI 


m . Tjy^ 
'' 2EF 


because 


m 


• S 
A 







Now we have previously shown that due to the shear there 
is a change of angle equal to -p— -7 Vf 

. • . Total change due to both causes 
_m ( ay , rjy^ 



k^\hG ' 2E 

_ m / ay r] y^ G 

^ GkA b'^ 2E 

^ 0" 1 771 / a tj ij \ 

putting E = and v = a ^-^^^ comes to ^-r^ ( -, - "^ 90 ) 

From this relation the slope at any portion of the section 
can be found, and the distorted form of the cross section can 
be obtained. Our present scope prevents us from dealing 
with this interesting problem further, but what we have given 
should serve as an indication of the method in which the 
problem may be attacked. 



CHAPTER XVII 

* FLAT PLATES AND SLABS 

In the beams that have been considered up to the present 
there is a support along two edges only, that is the support 
is at most along two parallel lines; when a plate or slab is 
supported upon more than two straight edges we have to 
consider the strengthening effect of the side supports, and for 
this purpose slab formulae are required. 

CIRCULAR PLATES AND SLABS 

Slab Coefficients. — In many problems it is convenient 
to use slab coefficients to compare the bending moments in 
a slab supported on its edges with the corresponding cases 
in which the slab is supported on two edges only as in the 
ordinary beam. We then have 

/? = slab coefficient 

B.M. on slab 

~ B.M. on corresponding beam resting upon two edges 

Bach's Theory. — Bach obtains formulae in a very simple 
manner by assuming the supporting pressure imiformly dis- 
tributed along the edge of the plate or slab and calculating 
the bending moment over various sections. In the use of 
these results it should be remembered that they give the 
mean bending stress across what corresponds to the breadth 
of the beam, but that they do not give the absolute maximum. 
A similar point occurs in considering shear stresses in beams 
(see p. 473). 

We will take the following standard cases. 

A. Circular Slabs supported on the Outside Edge. — 

488 



FLAT PLATES AND SLABS 



489 



(1) Uniformly Distributed Load W. — The reaction pressure 

W W 

per unit length will be P = ^ -n ^ o — t>- Then considering 

TT U Z TT Jtv 

w 

the forces in one half of the slab we have a load ^ acting 

downwards at G^ (Fig. 232), the " load-centre " or centroid of 

W 

the semi-circular area and a resultant reaction = -^ acting 

at the " reaction-centre " G„ or centroid of the semi-circular 



arc. 





W 



onocxiooooooooo 



Fig. 232. — Circular Slab supported at the Edge with Uniform Load. 

. • . Taking moments about x x we have 

W 

Bending moment on x x = M^ = -^ . o G^, — Wg o G^ 



2 
4R 



(OG„- OG,) 



^^. , 2 R 

o C}^ = ^ — and o Gj, = 

O TT TT 



. O G, 



- O G, 

WR 

w; TT R2 . R 



R / 4\ _ 2 R 
^ ~ SJ ~ 3 tt" 



or, if w is the load intensity of load — 



3 



(1) 



490 THE STRENGTH OF MATERIALS 

The mean stress on x x = / = -^^ = .g^ = ^-^^ 

. ••/ = ",?' (2) 

If the slab were freely supported at y and y we should have 
the reaction acting at y and the load at g, . 

- -288 W R 

.• . Slab coefficient = /3 = ^^ -^ 288 W R 

= -368 
(2) Central Load on Radius r. — In" this case as before we 

have 

^ W 

B.M. on X X (Fig. 233) = M, = -^ (o g,. - o gJ 

_ W/2 R _ 4r 

2 \ TT 3 TT 

-'=s-t;r('-ia i« 

In the limiting case where r = we have the point load for 
which 

'='j i») 

In this case if the supports were at y y we should have 

_WR/ ir 

~ 2 V 37rR 

2(1-^^''- 

^=T'-4?^^ ^^^ 

\ SttR. 

= "636 for the point load when r = 



FLAT PLATES AND SLABS 



491 



B. Circular Slab supported on a Circular Pillar. 

(1) Uniformly Distributed Load W. — In this case the 
tension and compression edges will be on opposite sides of 
those in the previous case, the present one being the equivalent 
of the uniformly loaded cantilever. 




OOOf)CY:CYYTrT^ 



w 

Supported at edge ; central load. Supported at centre ; uniform load. 

Fig. 233. Fig. 234. 

Circular Slabs. 



By moments as before about x x (Fig. 234) we have 

W W 

_ W /4R _ 2 r 

2 ^Stt 

_ WR/2 _ r 



If the load is w per unit area, W ^ wttVJ^ 

'2 r 



.'. M. =wW 



(7) 



(8) 



In the limiting case of r = 0, which corresponds to a point 
support, this gives 



492 



THE STRENGTH OF i\L\TERIALS 



Taking the corresponding beam we should have G^ acting 
on the edge of the supporting circle 



/3 = 



WR 


(5- 


77 r 


TT 


2R 


2 


r 




3 


R 




2 


irr 




3 ~ 


2R 






There is no slab coefficient corresponding to ;• = 0, or rather 
it would be more correct to saj' that it will be 1 in this case 

. /_ 6M _22rR-^ 
• • ^ ~2R.r^ ~ /2 i^) 

(2) Load Distributed uniformly along the Edge. — 
This case is the same as case A (2) with the loads and reactions 
reversed. 

C. Oval Plate supported on Edge (Fig. 235). — Load 
Uniform. — In this case we can obtain an approximate solution 
by assuming that the jioints G^ and g^ will be the same for 

Mj as for a circle of radius ^ and for M,, as for a circle of 



radius 



2 



FLAT PLATES AND SLABS 493 

6 TT 

6 M, W 6 
Wl 



This gives M^ = 

O TT 



^' It^ Tvlt^ 



O TT 

Similarly /, = -^ = ^^^, 
If we put W = — ^ — we have 

M -^fill" 
•" ~ 24 



M,= 


wb P 
24 






Corresponding 


to these 


we 


have 






/. 


w b^ 






h 





It will be noted that the stress is greatest across the short 
axis. This should not be used for ovals with Z ]> 2 6 which 
should be treated as ordinary beams of span b, giving 

M = ^ 

8 

Another way of dealing with this problem is as follows : 

If I is so great that the effect of the edges is negHgible, we have 

ID b t 

for the short span of a unit width B.M. = —^ and Z = -„ 

* • ^" ~ 8 ■ 6 ~ i^ 

For the circle, we have by Grashof 's theory (table on p. 495) 

_3dw'R^ _39wb^ _-3wb^ 
'' - 32 i^ - 1:287^ ~ ~^^ approx. 

. • . We may make up an empirical formula 
which is correct for the extreme values b — I and r = 



494 THE STRENGTH OF I\L4TERIALS 

Grashof's Theory. — Grashof investigated the strains 
in flat plates by an investigation of the deflected form and 
adopted the stress equivalent to the maximum strain as the 
criterion of the resultant stress (see p. 44), the result being 
different from that obtained by means of the calculation of 
the principal stresses. 

The derivation of the formulae is too complicated for our 
present scope, so that we will give the results only and refer 
the reader to Professor Morlej-'s Strength of Materials (Long- 
mans) for the mathematical deduction of the formulae. 



SQUARE AND RECTANGULAR SLABS 

Square and rectangular slabs are of importance in several 
cases in practical design, particular^ in reinforced concrete 
construction and in tanks and valve-chest covers. 

Rigorous methematical methods cannot be applied to these 
cases, so that we have to fall back on approximate methods of 
which the two following are the most common. These hold 
for uniformly distributed loads onh\ 

Grashof-Rankine Theory. — These formulae for slabs 
supported on their edges are obtained by the following reason- 
ing and should not be confused with the rigorous Grashof 
treatment for circular slabfe. 

Consider two narrow central strips (Fig. 236) of width x 
parallel to the axes x x and y y, thus forming one strip 
passing over the other so that the two strips must have the 
same deflection. The whole slab is considered as divided up 
into strips at right angles to each other, the strips passing 
one over the other. The load w per unit area may then be 
considered as divided up into two portions Wi, w,, carried 
respectively on the long and short spans. 

For the long span we have 8 = ^q-j-U t (1) 

,, short span we have S = '' ^ (2) 

oo4 hi i 



FLAT PLATES AND SLABS 



495 



< 

o 

w 
< 

Ph 

<1 
1^ 

u 

M 

O 
o 

CO 



O 

w 

cc 

o 






"SCO 
^1 



^fl 22 



Til 




p^ 




§ 


w 


o 


i;D 


00 


lO 


i-H 


(N 



q3 '^ 



^ 






r-H i"* 



o 






a 



c^ 



o 

p. 
a 

a2 



o 

Oh 






o 
&, 
P 



^1 
o o 



■ O 

CO 

^§ 

4J 1— I 
(D ,— , 

'+-' S-l 

-(J 

biD© 
^ O 



S 



-1-3 

© £3 
bc.aJ 



§'W 



Oi CO 






j3 














^^ 












^ ^ 


Stra 






(N 




M 


^|P^ 


S 






CO 


o 


t'pH K 


1 


;3 










-* Tt< 


1— 1 














Valu 
Maxii 


P5 - 


Ph « 


P^l^ 




+ 

P3!^ 
be 


11 


1—1 'rH 




o 

+ 


be '" 

O 4^ 
— 1 M 

^- — ^ s 


ot 






Ti4 !lO 


'^ Ico 


o^ 


■ess equi 






^ .CO 


IN 

Ph ^ 


^ 










»o ^^ 


02 










^ ; 








« !^ 




^i^ 








5* 




IN 










^,a 




P^ ^ 


1 


02 










T+< . 


I-H 


OQ 






1 


1—1 




incipal Stre 


IM 

Ph «,. 

CO CO 


IN , 

p^ «, 

CO \-^ 


P^|5^ 
«) 

bO 

+ 


be 4^ 

O M 


CO ,Ti< 

+ 

p^l^ 

bJO 


Ph 






Th !io 


f^ (M S 

I-H 00 


^ 








^ 










lO ^ 


j 


f^l-. 








—1 loo 




S loo 










1 
















CO ' 



C =H-I 

o o 

GO 



T3S 



II c3 



o o 



THE STRENGTH OF MATERIALS 

w,, l^ 



496 

These must be equal .*. 



also iVi, -^ Wi = w 
i.e. Will + , 4 j = w 



Wt = 



w 



1 + 



Y 



vC- 



"^ 



Y 



Fig. 236. — Grashof-Rankine Theory of Rectangular Slabs. 

.'. M^ = B.M. on short section, i.e. B.M. on long span. 

'w, P- w P 



1 + 



(3) 



Similarly w,, 



w 



1 + 



M, = B.M. on short section, i.e. B.M. on long span. 
_ w,, b^ _ IV b^ 



1 + 



6* 



'. For long span, i. e. short axis — 

B.M. on slab _ ^ b^ 
^' ~ B.M. on corresponding beam ^* + 6* 
For short span, i. e. long section 



I* + h*' 



(4) 
(6) 



FLAT PLATES AND SLABS 



497 



Some confusion is likely to arise if we do not clearly keep in 
mind the fact that if we are considering the stresses across the 
long section we take the B.M. on the short span, 

/• I M I n n 1 1 1 1 m 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n H I 1 1 1 1 n 1 1 1 1 1 1 1 1 n 1 1 m •<? 



-a 



I 



t 



I 



v< 



ii: 



.1:. 



W // L2 (3 



/.'9 (£ /£ 1.7 

Length -^ Breadth. 
Fig. 237. 



V'l: 



a.^ 



m 19 zo zi 



As a check we may remember that the greater B.M. always 
comes on the short span. 



I 

b 


Slab Coefficients 


Short Section Long Section 




and Long Span and Short Span 




Pi pb 


1 


•500 


•500 


1-25 


•291 


•709 


1-5 


•164 


•836 


1-75 


•096 


•904 


2 


•059 


•941 



KK 



498 THE STRENGTH OF MATERIALS 

Intermediate values may be obtained from Fig. 237. 
In comparing this theory ^^ith Bachs it should be remem- 
bered that the present gives the maximum stress, whereas 
Bach's gives the mean stress across any section. 

Xu^iERiCAL Exa:mple. — ^4 rectangular slab 18 ft. long and 
12 ft. wids carries a uniformly distributed load of 200 lbs. per 
sq. ft. What B.M. should it be designed for across the loyig and 
short section respectively ? 

In this case W = 200 x 18 x 12 

.'. B.M. on long span, neglecting slab action 

_ WJ_ _ (200 X 18 X 12) X (18 x 12) 
~ 8 ~ 8 

= 1,166,400 in. lbs. 
B.M. on short span, neglecting slab action 

Wb (200 X 18 X 12) X (12 x 12) 







8 


= 


777,600 in. 


8 

lbs. 




From 


our 


table 


for 


6 = !•- ^ 


= -164, and (i, = 


•836 




. 


. B.M 


. on long span 


= -164 X 1,166,400 






B.M. 


on 


short span 


= 191.000 in. lb<. 


nearly 




= -836 X 777,600 














= 650.000 in. lbs. 


nearly 



. • . On short span if / = 16,000 for a metal slab we have for 
short span 

16.000 . ^^ X 12 = 650.000 

D 

_ 6.50.000 
~ 576,000 

= 113 

t = \ TT3 = H" nearly. 

The other span would require a smaller thickness. 

In metal slabs the modulus of section is probably the same 
in both directions, but in reinforced concrete slabs the modulus 
generally varies. 



FLAT PLATES AND SLABS 



499 



Bach's Theory = — The pressure is taken as uniform along 
the edges. 

(1) Diagonal Section. — Assuming that the diagonal sec- 
tions are the weakest because tests indicate that failure often 
occurs diagonally, consider the bending moment about the 
line AC (Fig. 238). 

Let p be the pressure per unit length along the supports 




Fig. 238. — Bach's Theory of Rectangular Slabs. 

and let W be the total uniformly distributed load and w its 
intensity per unit area. 



Then 



V 



W 



(i; 



2 (Z + 6) 

The supporting forces or reactions may be taken as a force 
equal to "p b acting at Y and one equal to pi acting at x ; the 

W 

load on the A a b c = ^ and acts at the centroid G of the A. 

The perpendicular distance of x and Y from a c are each ^ 



and the perpendicular distance of G from a c is ^ 



500 THE STRENGTH OF MATERIALS 

. • . Taking moments about A c we have 

Bendmg moment = M^, ^2 2 2*3 






a fW _W^ 
2 12^ 3 / 
Wa 
12 



(2) 



But a X A c = 2 area of A a b c = Z 6. 

lb 
.' . a = — > 

AC 

.. M =^^^^ 

'' 12 AC 

W/6 



12 ^/l^ + 62 



(3) 



Neglecting the support on the short side we should have 
• ^ M,^^^ (4) 

To get a reasonable comparison between M^^ and M^. we 

ought to compare the B.M.s on the same length because, of 

course, a c is greater than Y Y. 

M M 

. • . B.M. per unit length along a c for slab = — ^ = --=^^ 

WZ6 



i2 (Z2 + 62) 



(5) 



W6 wb"^ 
B.M. per unit length along Y Y for beam = -^,- = — g- 

Diagonal slab coefficient 

- f^^^ - 12 (^2 _|_ 62) • 8^ 3 (^2 + 62) 

2 



3(1 +(f^ 



(6 



FLAT PLATES AND SLABS 
p has the following values — 



501 



I 


Diagonal 


b 


Slab Coefficient. 


1 


•333 


1-25 


•407 


1-5 


•461 


1-75 


•502 


2 


•533 



To use these figures we find 



8 



and treat that when 



multiplied by /? as the mean B.M. per unit length along the 
diagonal. 

Numerical Example. — Take the same case as worked out 
on p. 498, and adopting a stress of 16,000 lbs. per sq. in. find the 
necessary thickness of a rectangular metal slab, comparing the 
results by the two formulce. 

On Bach's Theory — 

— ^ = Q = 3,600 ft. lbs. per foot width 

o o 

= 3,600 in. lbs. per inch width. 

.'. B.M. per inch length on diagonal = '461 x 3,600 

= 1,660 in. lbs. nearly. 



M 



/ ^ = 16,000 X 1 X ^^ 
' ' 6 ~ 6 



2 _ 



6 X 1,660 



= -62 



16,000 

t = •\/-62 = -79 in. nearly, say y^ 

(2) Sections Parallel to Sides. — The following modifica- 
tion of Bach's treatment is more suitable for reinforced 
concrete slabs where the reinforcement is parallel to the edges. 

We can consider in a similar manner the strength of the 
section x x (Fig. 239). 

The reactions on the sides will have reactions at the mid- 

points equal to ^ at d and E and pb &tY. The load acts at 
the point r. 



502 



THE STRENGTH OF 1VL4TERIALS 



Therefore, taking moments about the hne x x we have 

pi I ^ I \N I 

2 4 ' ^ 2 2 4 

WZ 



= |'(26+/) 

_ W Z (2 6-fJ) _ W I 
~ 8 (r+ 6) 8 

Wbl 



S{1 ^b) 



Y 



± 



B 



(7) 



Fig. 239. — Rectangular Slab : Modified Bach Treatment. 



Neglecting side support on long side, we should have 

WJ 
8 

b 1 



M. 



.*. Slab coefficient for x x = ^/ == ,— , 7 = 7 

l+b ^^l 



(8) 



Similarly, if we consider the strength of the section y Y we 
should get 

Slab coefficient for y y = yS^ = 7 7 = z (^) 

^ +7 



FLAT PLATES AND SLABS 

These results can be tabulated as follows — 



503 



I 

b 


Slab Coefficients. 


Short Section 
and Long Span 
pi 


Long Section 
and Short Span 


1 


•500 


•500 


1-25 


•444 


•556 


1-5 


•400 


•600 


1-75 


•364 


•636 


2 


•333 


•667 



It follows from this that the B.M. comes the same on the 
two spans, so that the long span is the weaker as the breadth 
is less. Experiments do not bear this out. 

Numerical Example. — Taking the previous case we shall 
have /3i = -400, ft = •600 

.-. B.M. on long span = -4 x 1,166,400 

= 466,600 in. lbs. 
B.M. on short span = "6 x 777,600 

= 466,600 in. lbs. 

iMf = 466,600 
6 

466,600 



on long span 16,000 x 



P 



= 1-22 



384,000 

t = V'l'22 = 1| ins, nearly. 

This agrees fairly well with the Rankine value, although it 
is deduced from a different span. It is interesting to note 
that although Bach worked upon the diagonal as being the 
weakest section the above treatment requires a greater 
thickness. 

Variation of Bending" Moment. — To obtain an idea of 
the variation of the Bending moment in this case consider a 
portion a b u u (Fig. 240) of the slab. Then the load on 

I 

W 

2{l + h) 



the shaded area =^ 
As before, we have p 



504 



THE STRENGTH OF MATERIALS 



. We have at x a force p h and at d and e a force ^^, and 

W:r 



at F a downward force 



I 



. • . Taking moments about u u Ave have 



M,=phx +2'P^ .^ 



= Wx 



21 



b X x^ 

2 {I + b) "^ 2 {I + 6)"~ 2l[ 

Wr 

{bl + xl -x{l + b)} 



2l{l + b) 

^^ fhJ -h X _^bx{l-x) 
27(^ + 5)^ ^^ ~ 2l{l +b) 



Y 



d 



U D A 



2 



Y 
Fig. 240. 




putting X = ^ this gives 



(10) 



M^ = oi^j—TTx = o /7 , rx ^s beiore, thus checkmer 
8 Z (/; + 6) 8 (? + 6) ' ^ 

W6 

^" = 21(1+6) i^^-^'> • (11) 

A diagram showing the variation of M^ with y would come 
a parabola with vertex at the centre, similar to the ordinary 
B.M. diagram for a simply supported beam. For a corre- 
sponding consideration for the other span we should get 

Wl 
Mj, = 6/h(f I ;.\ 1^ ^ "" ^^\ which would also be a parabola. 



FLAT PLATES AND SLABS 505 

We may therefore assume that the stress will vary across a 
section approximately in the form of a parabola, so that the 
maximum stress will be 1"5 times the mean stress ; this would 
require the thickness at the centre to be \/l'5 = 1"22, say 1"25 
times the thickness given by the ordinary treatment of Bach's 
Theory. 

In our example this would make t for diagonal consideration 
= "96, which agrees quite well with the Grashof-R-ankine theory. 

Variations of Bach's Theory. — There is reason to believe 
that the pressure in rectangular slabs is greater at the centres 
of the supporting edges than at the corners; we will there- 
fore consider two variations in pressure. 

Variation I. — Pressure varies according to a Parabola. — 
We will now therefore assume the pressure to vary in the 
form of a parabola as shown in Fig. 24L We will take, as 
before, the total pressure on each side proportional to its 
length, so that the total pressure on each long side 

= V ^ WZ 

' 2 [l + b) 
and that on each short side 

^P = W6 

' 2 (I + b) 

The pressure at the centre of each side is therefore 1*5 ^3, 
p being the value given in equation (1). The resultant pres- 
sure along A B will act at the point y, while that on the half 
sides A X, B X will be at the centroids of the parabolas, i. e. 

31 , 

,- irora X. 

Id 

Therefore, taking moments about x x we have 

M =P ^+?J^ ^-"^ ^ 
* ■ 2 ^ 2 ' 16 2*4 

Wbl 3WZ2 Wl 



4 (/ + 6) ' 32 (? + b) 
Wl 



[2b + ^^-{l + b)} 



S{1 + b) 

8ir+T)r 4j ^^^^ 



506 



THE STRENGTH OF MATERIALS 



Neglecting side support, we have as before 

M =WJ 



. • . Slab coefficient for x x = ^/ 



[l + b) 
I 



1 - 



4 6 



1 + 



/ 



(13) 




Fig. 241. — ^Rectangular Slab with Parabolic 
Distribution of Edge Pressure. 



Similarly, we get for y Y by reversing I and b 

b 



I 



Slab coefficient for y Y = (5i, = 



I + b 

I _ 1 

b 4 

1 + .^ 



(14) 



FLAT PLATES AND SLABS 

These results can be tabulated as follows — 



507 



I 

b 


Slab Coefficients. 


Short Section 

and Long Span 

|3/ 


Long Section 
and Short Span 

^6 


I—" 


•375 


•375 


1-25 


■306 


•444 


1-5 


•250 


•500 


1-75 


•205 


•545 


2 


•167 


•583 





4- 



Fig. 242. — Rectangular Slab with Triangular 
Distribution of Edge Pressure. 

Variation II. — Pressure varies according to a Triangle. — 
In this case we will assume the pressures to be even more 
concentrated at the centres than in the previous case, and 
assume the pressure distribution shown in Fig. 242. 

As before, we take total pressure on each long side = P,. = 

^^m , 7v and that on each short side = P, = 0/7 , i.\ 
2{l + h) 2 {I + b) 



508 



THE STRENGTH OF MATERIALS 



Taking the side pressures as acting as the centroids of the 
triangles, we get 

W I 
2 4 



M — P - -]- — '^ - 

^ • 2 2 '6 



'Whl Wl^ 

4 (r+ b) "^ 12 {I + b) 

Wl U. 2 1 



S{l + b)\^^ 3 



8 

(l + b) 



Wl 



S{1 + b)\ 3 
. • . Slab coefficient for x x = F/ 



(15) 



b - 



I 



I + b 

lJ ■ 

Similarly, by reversing I and b, slab coefficient for y y 

I _ 1 
b 3 



(16) 



= r, = 



/ 



(17) 



+ 1 



These results can be tabulated as follows- 



I 

b 


Slab Coefficients. 


Short Section 

and Long Span 

Pi 


Long Section 

and Short Span 

P" 


1 


•333 


•333 


1-25 


•259 


•407 


1-5 


•200 


•467 


1-75 


•151 


•515 


2 


•111 


•555 



Numerical Example. — J'aH?i^ the parabolic variation, 
calculate the thickness of the slab previously considered. 
We have, neglecting slab action — 

B.M. on long span = 1,166,400 in. lbs. 
B.M. on short span = 777,600 in. lbs. 



FLAT PLATES AND SLABS 509 

For I = 1-5, 13, = -250, B, - 500 



.-. M, = 388,800 ill. lbs. 
M.^ = 291,600 in. lbs. 

.-. Short span 16,000 x ^^ ^ = 388,800 

16,000x1^2x12^^^33^^^^ 

388,800 _ 
16,000 X 24 
t = Vl'Ol = 1 in. nearly. 

Rectangular Slabs Clamped on their Edges. — An 

approximate treatment commonly adopted in practice for 

W I W6 
this case is to regard the B.M.s at the edges to be - and -y^- 

respectively, multipHed by the slab coefficient derived for 

supported ends ; those at the centre to be — and reduced 

24 24 

in the same ratio. In cases where the load may be on only a 

part of the slab the B.M.s at the centre are usually taken as 

the same as for the edges. 



CHAPTER XVIII 

* THICK PIPES 

We have considered already the strength of a thm pipe and 
obtained simple formulae by assuming that the stress was 
constant throughout the length and thickness of the pipe. 
WHien a pipe is not very thin compared -^ith its diameter 
we have to allow for the variation of stress across the section. 

Lame's Theory.— Let a pipe be of internal radius r 
(Fig. 243) and external radius R and let it be under pressure 
either from the inside or from the outside. 

Xow consider an imaginary thin ring of thickness 8 x and 
internal radius x. This rmg will be subjected to a radial 
pressure p on the inside which hj considerations of sj'mmetry 
must be the same all round, and on the outside it will be 
subjected to a radial pressure which will differ sUghtly from 
y and which we ma}' call ;p -^ hp. This assumes that the 
tube is subjected to pressure on the outside ; if it is on the 
inside the same formulae hold with appropriate change of sign 
as explained later. 

We may therefore apply to this imaginary hoop the same 
treatment as for a thin pipe, the circumferential stress, or 
hoop stress, being /. 

Considering a unit length of pipe we have 

Force tending to cause collapse of ring 

= {p -{- Sp) X 2 {x + S X) 

Force resisting collapse of ring =^ 2 f 8 x ^ p x .2 x 

These must be equal 

510 



THICK PIPES 511 

. • . dividing by 2 and neglecting the product, S p .Sx,oi two 
very small quantities we have 

p.x-{-x.Sp-\-pSx ^ f Sx + p . X 
.' . {f — p) h X = X h p 

(/ - ^) = T^ 




Fig. 243. — Stresses in Thick Pipes. 

In the limit when the increments are infinitely small this 
gives 

if-p) = ''ff w 

This is one relation between / and p. 

Now let us assume that the strains along the length of the 
pipe will be such that a plane section before subjection to 
pressure remains plane after subjection to pressure, i.e. that 
Jongitudinal strain is constant. 

i. e. '^J^ + "^-J = constant (2) 



512 THE STRENGTH OF MATERIALS 

because both / and p will cause transverse strains in the 
direction of the length of the pij)e, and they will have the 
same sign. 

. ■ . Since -q and E are constant, if our stresses are within the 
elastic limit we may write 

f + p — constant = 2 a (say) 

••• f = {2a-p) (3) 

Put this value in (1) and we get 

2o X d p 

a — 2 p =^ , - 
ax 

2a = 2p + ='^J (4) 



Cu tX) (aj JO 



= x( 2 2> + 1 \ = 2ax 
. • . d {px^) = 2 ax . dx (5) 

Integrating we get 

p x^ = a x^ +6 
where 6 is a constant 

-'• P -CL + ^2 ••• (6) 

but / = 2 a - 2> (by 3) 
i.e. f =^a --^ (7) 

by calculating a and h in any particular case we can find 

formula for p and /. 

Special Gases. — (1) Pressure Inside = pi\ Pressure 

Outside = 

i. e. p = Pi for X = r 

p = for a; = R 
••.^. = « + J W 

b 



THICK PIPES 513 

Put this value in (8), then 

■■■^-^ (9) 

. • . Hoop stress at inside — /,• is obtained b}^ putting 
a; = r in (7) 

I.e. /. =a---2 

_ fi r^ Pi R2 

- - R2^ir^2 - R2 _ ^2 

~ R2-r2 ^^^^ 

The negative sign indicates that the stress is a tension. 

Hoop stress at outside = fo = a — p^ 

= - (R2 _ r^) ~ (R2~_r72) 

This is also a tensile stress and is clearly less than /,, so that 
with internal pressure the maximum stress occurs on the 
inside. 

At any intermediate radius x 

, h 

— Pj ^^ _L P' "^^ ^^ 

~ ~ (R2""^2j + (R2 _ ^2) ^,2 

- ^'-^^ r5!_i\ n.N 

(R2 _^2) \^2 -^j U'^; 

It should be noted from equation (11) that no matter how 

LL 



514 



THE STRENGTH OF MATERIALS 



great the thickness of the tube may be the hoop stress is 
always greater than the internal pressure, so that for any 
given material there is a certain maximum pressure which 
must not be exceeded. 

It should also be noted that the assumption of the constancy 
of longitudinal strain holds only while the stress is within the 
elastic limit. 

Numerical Examples. — (1) A cast-steel cylinder 2 ft. in 
external diameter and 3 inches thick is subjected to an internal 
pressure of 2 tons per s(l- in. Where and of what magnitude 
is the m^aximum stress ? 

The maximum stress is on the inside and is given by the 
formula 

p (R2 + r^) 



ti 



(R2 - 

2(242 



r2) 
f- 182) 



(242 _ 182) 
'7 



2 X 62 (16 +^) 
62(16 -9) 



7*14 tons per sq. in. 



(2) Plot a curve showing the maximum stress in terms of the 
internal pressure in a tube whose ratio of external to internal 
radius varies from TIO to 4. 

_ p, (R2 + r2) 

U (R2_r2) 

/, _ R2 + ^2 

' Pi 



R2 - r 

?)• 

R^2 

r 



+ 1 



This gives the following values- 



R 

r 


MO 


1-20 


1-30 


1-5 


2 00 


2-50 


300 


3-50 


400 


f' 
Pi 


10-52 


5-55 


3-90 


2-60 


1-67 


1-38 


1-25 


118 


113 



THICK PIPES 



515 



If- = 110,^^-^ = 110, 
r r 



10. 



T) V f 

. ' . The thin pipe formula / = ^~~- would srive — = 10, so 

t p 

that the thin pipe formula would be about 5 % in error. 

The above figures give the curve shown in Fig. 244. These 

results should be compared with Example 2, p. 521. 



10 
9 
8 
7 
6 
S 
4 
3 
Z 
































































































































































1 


I 


























\ 


























"^ 


\^ 














































1 







































Valuea of R -^ r 



z-5 



3'0 



3-5 



4o 



Fig. 244. — Variation of Hoop Stress for various ratios of External 
to Internal Radii of Pipes with Internal Pressure. 



Curves of Variation of Radial and Hoop Stress for 

Internal Pressure for R == 2 r. 

T3 ^. ,^, b RV2 4r4 4r2 

By equation (9) ^- = ^,— ^ = 37^ = ^ 



By equation (10) ^ = - ^2 

Pi Pi PiX^ 



516 



THE STRENGTH OF MATERIALS 





1 4 ^2 
~ 3 "^ 3 .1-2 




-sKx) 3~3rv.i-7 / 


/ 


h 

= ^-7^ 


L 


(1 4/rYl li.fr 



These results are plotted in Fig. 245 and show clearly the 
variation of the stresses across the section. 

10 



•47 



1 t 



\ 




















\ 




















\ 


\ 




















\ 


















\ 


















... . 


N 


\ 




















\ 


\, 




















\ 


k, 






















\ 


"^ 






- 














'^ 


^ 


^ 



1-17 



ro7 



•67 



K 



b7 > 



1 + 



1-8 



10 



Values of 2 4- r 
Fig. 245. — Variation of Stress in a Thick Pipe with 
Internal Pressure. 

Maximum Shear Stress.— The stresses / and p are the 
principal stresses in the material and, as we showed on p. ID 
the maximum shear stress will be equal to 

s = ^-^ = ^,, (from (6) and (7)) 

^ X" 

p, R^r2_ 

~ (R2 - r2).T2 

The maximum shear stress will occur Allien x has its least 
value, i. e. at the inside edge where x — r 



(15) 



THICK PIPES 



517 



Max. shear stress = 



p^B,^ 



(R2 - r2) 
Maximum Stress equivalent to Strain. 



a strain in the circumferential direction equal to — ^ 

. * . Total circumferential strain = 4^ — %^ 

E E 

. • . Equivalent stress = Total strain x E 

= fe = f-VP 

b (1 + 7}) 



(16) 

There will be 

IP 



= a{l -rj) 



x" 



(17) 



This gives a maximum equivalent stress at the inside of 

p, (R2 + ^2) 



fe 



(R2 - r2) 

/RM-^2 

'^ 1112 _ ^ 



y'i -P2 _ ^2 + vj 






(18) 



putting y} = \ this gives 

/. = - Vi 



R2 + r^ 1) 

R2 _ ^2 + 4j 



NuMERiCAL Example. — Take the case of a tube with internal 
radius 4 in. and external radius 12 in. and take p = 10,000. 

This case was given by Professor C. A. M. Smith,* and we 
have added the equivalent stresses. 

This gives 6 = 180,000, a = - 1,250. 

The results may then be tabulated as follows — 



Radius. 


Maximum Stresses. 






I 


1 Hoop Tension 


Compression. 


Hoop Tension. 


Shear. 


equivalent to 




P 


/ 


s 


Strain. 


4 


10,000 


12,500 


11,250 


... 
15,000 


5 


5,950 


8,450 


7,200 


10,950 


6 


3,750 


6,250 


5,000 


8,750 


7 


2,420 


4,920 


3,670 


7,420 


8 


1,560 


4,060 


2,810 


6,560 


9 


970 


3,470 


2,220 


5,970 


10 


550 


3,050 


1,800 


5,550 


11 


240 


2,740 


1,490 


5,240 


12 


1 

1 


2,500 


1,250 


5,000 



* Engineering, September 2, 1910. 



518 



THE STRENGTH OF 1VL4TERIALS 



Formula for Thickness of Pipe in Terms of Internal 
Diameter and Pressure. — On the maximum stress theory we 
have, if ft is the safe tensile stress, 

_ p, (R^ + r^) 
'' R"2-r2 

.ft R2 + r2 



1. e. 



U + Vi 

ft - Vi 

R 

r 

r + t 



R!» - 
2R2 

2r2 



U±Vi 
U 



Vi 



t 



ft + Vi 



r r y ft — Vi 

t Ift+Vi 1 



--'{^'ikr') 



tt+Vi il 



ft - Vi 
On the Maximum Strain Theory we have 



, /R2 + r2 \ 



h 

Vi 



V 



A + (1 - 


- -n) Vi 


A - (1 + >?) p, 


//. + (1- 


- v) Vi 



A- (1 



v) Vi 

.'. t ^ r 

t =r -\ 

If q = -3 

t = r 

v 
t 



^ R^ + r^ 
~ R2 - 7-2 

_ R2 



(V!:^ 



r 

+ (1 



(19) 



If -. = i 



-(!+>?) V, 
4/7+3y. \ 

A - 1-3 p, 



- 1 



- 1 



- ' KV 3 /! 



3 /, + 2 p, 
4p, 



- 1 



..(19a) 
..(196) 
..(19c) 



THICK PIPES 519 

Fig. 246 shows a chart reproduced from Machinery, 
January 14, 1915, for determining the necessary thickness of 
a cyHnder or pipe in accordance with Formula 19. 

In using the chart to determine the constant, the horizontal 
line through the proper pressure value is located, and the 
curve starting from the desired value of the stress is next 
found; this curve is then followed to the point where it 
intersects the horizontal pressure line, after which the vertical 
line is followed to the bottom of the chart to determine the 
constant. This constant multiplied by the inside radius r of 
the cylinder gives the required thickness of the cylinder 
wall. 

Case 2. — Outside Pressure = %, Inside Pressure = 

In this case p = p^ when x = R 
and p ^ when x — r 
b 



I.e. 



Po 


= 


a + ^^ 





= 


, b 


a 


== 


b 

r2 


Po 


= 





P» - P"^'''' (20) 



J^ _ 1 \ (R2 - r^) 

R2 rV 

a = ^Jt^:^ (21) 



(R2 - r2) 

Hoop Stress at Inside. 
At inside, where x = r, fi = a 



b 



(R2 _ ^2) + (J^2 _ ^2) 

2 Po R /Qnv 

(R2 -r2) • ^^"^^ 



520 



THE STRENGTH OF :\L\TERIALS 



1 


STRESS IN CYLINDER WALL IN POUNDS PER SQUARE INCH. 


PRESSURE IN CYLINDER IN POUNDS PER SQUARE INCH. 


— 1 — 


























\a 


/ 








J 


A 


< 






























/ 






^ 


^ 




1 

_4 




























/ 




1 


A 
































y 


V\ 


.A 


X 






^ 


^^ 


^ 




> 


















. 


^ 


/ 


A 


^^ 


^ 


^ 


.^ 


^ 








- 
















/^ '>^ 






-=i 


li^ 


1! 














/ 




y^^ i^ 


^-J 


^ 


1 >--r-' 












1 


\ 


/ / 




X 1 


/I 


■^ J 

1 j-^ 


>^^' 












// V 


A\ 


a-:a. 


A 


i J 


X^ 










ll\U'l±^ 


X'-X^^\yA, 




\ \ \ 1 / 




^^ 


\>^ 


"! i 


1 1 // /// Y X y^ 


^ 


-^ li- 


-f^ 


\ 1 1 //// /y^y\^^ 


X 


^ 


M 


1 ! 


^i,^ 




^Ihf/l/y/ '/ 


/ 


A 


/ 






^ 


^ 


y 










'illll//'// . 


/ 






y 


y 


/ 












i i/////7// /' / 






/ 


^ 










^ 


X^ 


\ ///////;/ / /\ 


/ 


/ 


^ 






X 


^ 


r" 








/ // // \/ ■ / 






^/i 
















\ /////(/ 




/fi 


A 












.^ 




-^ 


\ ///// /\/ 


Y 




\x 








^ 


^ 


^ 


^ 












'/////// 


/ 




^ 


^ 


^ 




















III//////X 


y 


V 














_^ 


.^ 


H^ 


fT 

1 


11 //////y 




/ 


y^ 








^ 


- 




"^ 








i 


_ 


llli/// 


r- 

V 


<y 


y 




1 — i 


^ 










^ 


■^ 






1 1 




i 


w//y\>^ 


^i^^'^n^ 






^ 




— 


— 


—- ' 




r 


















_^ 








— 


— 


i 








\ 




















S ^ ^ <>t,<>i ^. ^. ^. ^. '^. "^ '-■'. '-''. ® ^. R ^. *"•. =°. °°. ^. ^. '^. '- 
0* 0* s' © 0" 1- 


chinery 



Fig. 246.— Diagram for Thick Tubes. 



THICK PIPES 521 

Hoop Stress at Outside. 

At the outside, where x = R, hoop stress = /„ 

_ _ b 
- ct ^2 

-^^^' + ''^ (23) 

In this case the maximum hoop stress, which is compressive 
throughout, also occurs on the inside and is greater than the 
radial pressure ; irrespective of the thickness of the pipe, the 
external pressure may never be more than half the safe 
compressive stress on the material. 

Putting in the values of a and b in the general formula for 
this case we have at a radius x 

_ p„ R2 p„ R2 r^ 

^ ^ (R2 - r2) - '(lR'2'372y ^2 

P"^^ A-fYj (24) 



(R2 - r2) I \x 

' = iBp^, !■ - ©■} « 

Numerical Examples. — (1) A cast-steel cylinder 2 ft. 
internal diameter and 3 in. thick is subjected to an external 
pressure of 2 tons per sq. in. Where and of what magnitude 
is the maximum stress? {Compare Example 1, p. 614.) The 
maximum stress occurs on the inside and is given by the formula 

- 2 po R' _ 4 X 24^ _ 4 X 42 X 6^ 
I' - (1^2 „ ^2) - 242 - 182 ~ 62(16 - 9) 

4 X 16 

= — ^ — = 9' 14 tons per sq. in. 

(2) Plot a curve showing the maximum stress in terms of the 
external pressure in a tube whose ratio of external to internal 
radius varies from 110 to 4. 

2^^ 

p, ~ R2 - 7-2 ~ f-Ry' 

r 



522 THE STRENGTH OF IVIATERIALS 

This gives the following values — 



R 

r 


110 

1 


1-20 


1-30 


loO 
3-60 


2 00 


2-50 


3-CO 


3-50 
218 


4 00 




1 

11-52 


6-55 


4-93 


2-67 


2-38 


2-25 


213 



R R 

For = MO we have ^ = MO 

r R — ^ 



•. t = 



R 

ir 



IZ 


— 


























10 


— 





















































8 


— 


1 














. 


■ 








6 


- 


1 





























\ 






















a 






\ 






















•1-4 






\ 
















1 






\ 






















o 








'x,^ 




















«Q 








^^^^ 




















3 












. _^ 
















^^ 






























n \'5 

Values of R 4-r 



zo 



rs 



30 



35 



40 



Fig. 247. — Variation of Hoop Stress for various ratios of External 
to Internal Radii of Pipes with External Pressure. 



The thin pipe formula would give 



= 11 



SO that the thin pij)e formula would be about 5 % in error. 

The above results gives the curve shown in Fig. 247, which 
should be compared with that shown in Fig. 244 for internal 
pressure. 



THICK PIPES 



523 



Curves of Variation of Radial and Hoop Stress for 
External Pressure for R = 2 r. 

h _ _ R^r ^ _ _ iL^ 

r2 - 3 



By equation (20) 



Vo 



R2 



. £ 

* ' Vo 


R2 


4 


R2 - r^ 


3 

'^1 



Vo 



vo 




















■^ 


"8 














^ 


^ 














,/ 


^ 










•6 

•1- 








/ 


/ 
















/ 


/ 


















/ 


















/ 


















'3 

P^ -2- 
o 


^ 


/ 


















/ 




















> 


/ 





















1-4 



1-8 



Vbl 



•S7 



Z'oy 



Z-27 



S5. 



02 



ft 

2-47 I 



VO 



rfo7 



Values olx-^r 
Fig. 248. — ^Variation of Stress in a Thick Pipe with External Pressure. 





4 4r2 




~ 3 3^2 




_4/ /ryi 


1 

Vo 


= («-|.)-2^- 




4 4r2 
~ 3 ^ 3 a;2 



(26) 



=J{'+(iT) ™ 

These give the curves shown in Fig. 248, which should be 
compared with Fig. 245, 



524 THE STRENGTH OF MATERIALS 

Strengthening Thick Pipes for Internal Pressure by 
Initial Compression. 

We have seen that the maximum hoop stress is always 
greater than the internal pressure, so that if we are to be able 
to make pipes sustain very high pressures we must devise 
some method of reducing the hoop stresses. This may be 
effected by bringing the metal of the tube into a state of 
initial compression. 

In the early days guns were cast around chills to cause the 
metal to solidify immediately on the inside and so come into 
compression when the remainder of the metal contracted 
upon cooling. 

Another method, now commonly adopted, is to wind strong 
steel wire under heavy tension on the outside of a tube, thus 
bringing it into compression which will balance to some 
extent the pressure caused inside the gun by the explosion. 

A further method is to shrink an outer tube on to an inner 
one ; this puts initial tension stresses into the outer tube and 
initial compression stresses into the inner tube. The effect 
of the shrinking is shown in Fig. 249. The top diagram 
indicates the distribution of the hoop stresses across the 
section for a solid tube ; the shrinkage stresses are shown in 
the central diagram ; these being obtained by applying the 
condition that the radial pressures of the junction are equal 
and opposite. 

The combined stresses are shownin the bottom diagram, from 
which it is seen that the maximum tensile stress is very much 
reduced and that the tensile stresses are more nearly constant. 

Necessary Difference in Radius for Shrinkage. — 
For the outer and inner tubes, the same general formulae will 
hold, but the constants will be different. 

i.e. For the outer tube we have 

p = a, + ^ (1) 

f =a.-k (2) 



THICK PIPES 



525 



For the inner tube 



V ^(^i+ ~2 



(3) 





ens/on 



Shr/'nnaqe 
Stresses 



I M I I I i I I I I M I 
Combress/on 




I 



Stresses 




y/Uulhide Tuhe'^nsidc TuLe\^ 



R 



-^ 



-H 



Fig. 249. — Shrinkage. Stresses in Compound Tube. 



f ^ Cli - 



x" 



(4) 



526 THE STRENGTH OF MATERIALS 

We must have the same value of p for the junction where 

• n A. ^" — 1 ^' 

i. e. {a,. — a^) r^ = (6, — 6,.) (5) 

Next consider the circumferential strains at the junction. 
For the outer tube we have 

Unital circumferential strain = != -\-^^ (6) 

Similarly for inner tube 

Unital circumferential strain = — (^ + ~f j . . .(7) 

The value of y is the same in each case, and in (6) / is as in 
(2) and in (7) as in (4). 

.*. Increase in circumference of outer tube 

2 7rri 



4(-'j 



+ >7P 



E 

Decrease in circumference of inner tube 

.". Difference in circumference of two tubes before heating 
and shrinking on 

. • . Corresponding difference in radius 
_ difference in circumference 

2 r 
= (from (5)) -^ (a, - aj 

2 
i.e. Proportional difference in radius = -p, (a^ — a^). . . . (8) 

This will probably be made more clear by the following 
numerical example. 



THICK PIPES 527 

Numerical Example. — A compound mild-steel cylinder con- 
sists of a tube 6 ins. in external radius and 4*5 internal radius 
shrunk on to another tube which has an internal radius of 3 ins. 
If the radial compression at the common surface is 4000 lbs. 
per sq. in. after shrinking find the circumferential stress at the 
inner and outer surfaces and at the common surface, and find 
also the original difference in radius necessary to effect the given 
radial pressure at the junction. 

Outer tube. 

p = for X ^ 6 

- 4000 for X - 4-5 
. n — _L -^ 

4h / 4 1 \ 

4000=..+ 3j^=^(,-^--3,)6. 

324 X 4000 

bo = - 



a„ = — 



7 
36,000 



X 



Inner tube. 



7 
Interior hoop stress {x = 4*5) 

_ _ 36,000 _ 324 X 4000 

~ 7" 7 '^ 81 

= 14,286 lbs, per sq. in. (tension) . 

Exterior hoop stress {x = 6) 

- _ 36,000 _ 324 X 4000 JL 
~ ~^7 " 7 ^36 

= 10,286 lbs, per sq. in. (tension) . 

p = for X =^ 3 
= 4000 for X = 4-5 

.-. = a, + g .-. a, = — 

4:b 
4000 = a, + ^ 

_ 4 6. _ 6, _ _ 5 b^ 

81 9^ 81 

,,= _400^2iAl._ 64,800 
o 

a, = 7,200 



528 THE STRENGTH OF ^UTERIALS 

.'. Interior hoop stress {x = 3) 

= 14.400 lbs, per sq. in. (compression). 
Exterior hoop stress (.r = 45) 

= 7.2W - : - ^-iii) 

= 10,400 lbs, per sq. in. (compression). 

Taking E = 30 x 10^ lbs. per sq. in. and v; = J. the cir- 
cumferential strain at 4 J ins. radius in the outer cylinder 

_ 14,286 , 4000 

~ 30 X 10« "^4 X 30 X W 
^^,^,.,^ 1000 

Circumferential compressive strain at 4 J ins. radius in the 

iiuier cvlinder 

_ 10,400 _ 4000 
~ 30 10^ 30 X 10« 

= ^103467 - 3,-^^-^e 

.• . original difference in diameter = 9 (0004762 -^ •0<X)3467) 

= 00741 in. nearly. 



CHAPTER XIX 

"^ CURVED BEAMS 

We have seen in Chapter VII that the ordinary formulae 
for beams hold only for cases in which the beam is initially 
practically straight. To obtain relations between the stresses 
and the bending moment in the general case we may proceed 
as follows — 

Let A B D E (Fig. 250) represent a short piece of a curved 
beam, o being the centre of curvature and a e and b d being 
sections normal to the centre line c &. Then, obviously, the 
material at E D will not require the same total strain to pro- 
duce a given unital strain and thus stress as the material in 
AB will, because its original length is less, and, as a result, 
the neutral axis will not pass through the centroid. 

While still making the assumption that stress and strain 
are proportional, and also Bernoulli's assumption that a 
section originally plane remains plane after bending, we can 
find a more accurate theory of bending of curved beams, as 
follov/s — 

Let the portion a b d e take up the position a^^ b^ d^ e^ after 
bending. Consider an element of area a situated at a point p 
at distance y from the centroid line c & and consider a fibre p q 
of the material enclosing the area a. 

After strain the fibre p Q takes up the position f^ q^ at 
distance y^ from the strained centroid line c^ c^' 



Then unital strain in p q = 



Pi Qi - p Q 



PQ 
MM 529 



)30 



THE STRENGTH OF MATERIALS 



And if fy is the stress at the point p 

/" _ Pi Qi - P Q _ Pi Qi 
E PQ PQ 

^E 



PQ 



Similarly unital strain along c c' = 



Ci c / — c c 
c c' 



and if /„ is the stress at the centroid, we get similarly 

u 

E 






(1) 



(2) 




Fig. 250. — Stresses in Curved Beams. 

Dividing (1) by (2), we get 

/ 1 + 4 
Pl Ql X c c __ E 

Ci C/ X P Q ~ " fo 

^ "^E 

Pi Qi _ Vi + Ri _ 1 , ^1 
Ci c/ Ri Ri 



But 



C Ci 



R 



1 



PQ 2/ + R . y 
^R 



CURVED BEAMS 631 

Also since !l and i^ are extremely small, we may write 

il_E ^ A _ /„ f, 

, , /o V' E + E 

1 + yy 

.-. We get 5! = 1 - ^ + ^' (3) 

1 + ^ 

^R 

1 + .a 
E E ^ y 



l + R 



1 _ ^1. 

= [y _L ^ ^1 

^R 

_^_ y 

]y Rj R 

E ^ t/ 
1 + - 
^ R 



E ''y^ y 



(4) 



/. - /o + ^^^ ^' (5) 

1 + ^ 
^R 

Then the load across the whole cross section is ^ f,j . a and 
in the case of pure bending this is zero. 

.-. We have 2^ . a = 

-^fyi y 



= 2 /. . a + S '-^^ ^ 



1 + ^ 

^R 



But ^ /„ a = /„ :§ a =: /„ A 

2/1 2/ 



.-./„= -^2^1-^. a (6) 



1+R 



532 THE STRENGTH OF MATERIALS 

The moment of the force on the given element about 
c c' — f,j . a ,y and the sum of these moments is equal to 
the moment of resistance and thus equal to the bending 
moment M 

. • . We have M = 2 /,.?/. a 

2/1 y 



^ VR R/ 

= :S /„ a . w + :S '^ — ^-— . a y 

^R 

But ^f„ay = f„%a.y— f„ X first moment of area about 
centroid = /„ x = 
. • . We have 



j^_ _^vRl ?V_^„ (7) 

(■+i) 

This is the most general case and is true for the assumption 
given. 

Now consider the following special cases — 

(1) Ordinary Straig-ht Beam ; R infinite, R^ very 

GREAT. 

In this case /„ = . 2 ,- * 

E 

Then Isl =^ ^ ^.y^y ex. 



y^ is practically equal to y 

_EI 
- Ri 

E II 
and from equation (5) /y = + -p 

_ E X 2/ 



CURVED BEAMS 533 

' ' ^1 y 

y 

This is the result we have previously obtained. 

(2) Winkler's Formula. — Winkler drew attention to the 
error of applying the ordinary bending formulae to chain 
links, etc., where the original curvature is appreciable, and 
improved such formulae as follows — 

He takes y\ = y- 

Then equation (5) becomes 

1 1 



/. = /.+^^^^^ ^ 



1 + ^ 
^R 

= A^+E(^-iV/T^, (8) 



.Ri r; > + R 

Then from equation (6) 

'" A VRi R/ V?/ + R 



I/ + R 
R.i/2 



Now let kh'^ = %, ^ 

.y + R 

where h is defined by the above relation, and may be called the 

link radius. It corresponds to the radius of gyration in the 

ordinary case. 

„, Ah'' ^f y"- 
.-.We see ^ = :S — ^-^ 

R \2/ + R 

^ 2/R 
2/ + R 

_y \ _ _Ah^ 
y ^-r)-"-- R2 

E /I l\Ah^ 



Th-/ =1.(^-1) 



A • VRi R/ R 



534 THE STRENGTH OF MATERIALS 

.•.Wehave/.,=^jJ':(^^-^) (9) 



From equation (7) 



M = :s ^^^ ^ 



1 1 



1 -^y 

R 



2/-tt 



ER(i--i^2 y'^ 



Ri r; R + 2/ 



Ri R/ " R + 2/ 

= E-^Ki-K) ■• ■•<^*^) 

returning to equation (8) we see 



R VRi R/ ' \R Ri; 2/ + R 

^ M M^ / R_y 

R. A^ A^2- VR +2// ^^ 

General Graphical Solution. — Let Fig. 251 represent 
the section a d b e of a beam, and o the centre of curvature 
of the centre line d e, the beam being, of course, curved in 
the plane of bending. 

Now consider a very narrow strip P Q of the half section at 
distance y from c D. Join p o, cutting c d in s and draw s R 
parallel to Q c to cut p q in r. 

^, p R R s y 

Then ^ 



p Q Q o R + 2/ 

Repeating this construction for a number of strips such 
as p Q, and joining up the points obtained, we get a curve 
A R D Ri B which is one-half of a curve called the link rigidity 
curve. In a symmetrical section, which is the most common, 
the two halves will be identical. 

* This is the stress due to bending only; in the case of hooks we 
have to add the direct stress over the whole section. 



CURVED BEAMS 
Then the area of this link rigidity curve 

= - A,* = -^ 



y 

R+2/ 



535 




Fig. 261. — Curved Beams, etc. 



■•■A„ = 
Now let -*- = 



R2 

area of half link rigidity curve 
area of half section 



= L i. e. A /^2 _ A X L X R2 

* The area is negative because the area of the portion d b^ b is greater 
than that of a p d . a,, represents the excess of the former area over 
the latter, i. e. area a r d Rj b — area a p d Pj^ d. 



536 THE STRENGTH OF MATERIALS 

Now put these values in the equation (11) for stress. Then 
we have 

'" ~ R. A + A.L;R2(R + y) 
' ^MM y _ 

A \R "^ R L (R + y) 

^ M f y I 

AR\' "^L(R+7)J ' 

Then if d, and dt are the distances from the hne D E to the 
extreme compression and tension fibres respectively, assum- 
ing the inside to be in tension and the outside to be in 
compression, we have 
Maximum compressive stress 

_ / _ M f d, 1 

Maximum tensile stress = ft = ir^^l r~-/^~ — yr ~ 1 < • • (1^) 

A R \^Li (R — dt) I 

Position of Neutral Axis. — The value of y to make /, = 
gives the distance y,, of the neutral axis from d e. 

'■ '• L (R + 2/.) " 

Vo = — L R — L t/„ 

LR R 



1 + L , R2 

This enables us to find the position of the neutral axis. 

Alternative Formula. — The stress formula on Winkler's 
assumption that yi = y can be put in a number of alternative 
forms. 

Suppose, for instance, that the neutral axis is at distance y^ 
below the centroid line c c-^ (Fig. 250). Then total strain at 
p Q = Pi Qi — p Q will be proportional to {y + y„) the distance 
from the N.A. 

. • . We write P^ Q^ — p q = m (?/ + y„). 

Moreover, p q = (?/ + R) x /e o d 



CURVED BEAMS 



537 



h _ JPiQi - P„9 _ (y + y.) m 
E P Q (?/ + R) * /e O D 



/. 



^ • V T^{ where ti is a constant 
(2/ + ^) 



(2/ + I^) 



(13) 




Fig. 252. 



Since 2 /„ . a = 



Eii 



y + y 

^ + R 



= 



1. e. y. 



y «■ 

^"+R 



a 

.V + 



1/ + R 



(14) 



538 THE STRENGTH OF MATERIALS 

Again, taking moments about the neutral axis we have 

Zj y y -^ Bo J 
. / = M (y + y,.) 

' ■ i»+«{2'i^"}- 

Resal's Construction. — ^According to this construction 
we proceed as before and find the link rigidity curve. 
The line n n passing through the centroid g of the curve 
A R Ri B r/ r' is then found by graphical or other methods, 
and the moment of inertia I^ of the link rigidity curve is then 
found about the line n n. 

In practice it is sufficient to apply the construction to 
one half only. 

Then N N is the neutral axis line and 

_MR {y + y„) 
^'!~ I, '(y + R) ^''^ 

The rule for the line n n passing through the centroid of 
the link rigidity curve is 





2/o 


%a^y 






Now 


aj 


a X R Q 


a 

R 


.R 

+ y 




yo 


^ <^y 

_Z/R + y 







ja^ +y 

This is the relation required by equation (14) 

Inn = S "1 (2/ + yo? 

/inN i MR (1/ + yo) 
.• . In equation (16) /, = i^Tj^I^y 



CURVED BEAMS 



539 



This formula looks simpler, but it involves the determination 
of i/o and I^, which are rather more troublesome than the 
calculations in the previous method. 

SPECIAL CASES 
(1) Rectangular Section. — If the section is a rectangular 



T 



-;-y 



dv 



-jt 



Fig. 253. 



one of breadth h and depth d, we may proceed by mathematical 
analysis as follows — 



A, = - 



b y 
R"T^ 



dy 



= -b 



- b 



/ 



1 - 



R 



n + y 



dy 



y-n log. {y + R) 
2n + d 



+ 2 

J d 
2 



-bid -n log. 



7 / 7 T-> 1 2 R -(- cZ 
b{ - d -\-R log, 2^ _ 



2n-d 
d 
d 



. L = 



K 
bd 



- ... 2R + d R, 



R 

d 



1 + 



A 

2R 



1 - 



R 



- 1 



540 



THE STRENGTH OF MATERIALS 



= |{log,.ll 

_Rr /^ 

~ d\ V2R 



_d 
2R 



)-log„(l 



J VI _i. 

2R/J ^ 



d^ , rf5 



(^6 



d-' 



A. 

2R 

(^ IR ' 

1 /^ 
12 VR 



2(2R)2 ' 3(2R)'' 4(2R)4 ' 5(2R)5 ' 6(2R)^ 7(2R)' 

_ d-^ _ ^3 d^_ _ rfs _ rfs _ d-' 

"'2(2R)2~3(2R)3 4(2R)^~o(2R)5 6(2R)^ 7(2R)- 

12 VR; ^ 80 VR/ "^ 448 VR/ / 
2 l.fd\^ . 1 /^\^ 



r> + 



80 VR 



+ 



448 



(if- 



(2) Circular Section. 



Fig. 254. 



A, = - 



h y dy 
t-{^y 



+ r 



-r -r 

-r + r 

"Rhdy 



"^kbdy 



-r 

+ r 



R + 2/ 



R-fi/ 
h dy .. 



-f r 



The second term / h dy = area of circle = ttt- 



}- 




(1) 



* See, for instance, C. Smith's Treatise on Algebra (Macmillan), p. 383. 



CURVED BEAMS 



541 



R 



+ 9 



, r h dy _ -p /"2 r cos d {r sin 0) 
1/ R+^~ y R+TsiiT^^ 



+ ■ 



= 2r2R 



/ 



C0S2 ^ ^ ^ 



R + r sin 61 



(2) 



_ cos2 e _ (l-si n^^) _ 1 __^( • 0_ I^sin ^ 

R4-rsin6' R+ r sin ^ ~ R + rsin(9 r\ R + r sin ^ 

1 sin ^ R / sin 



R + r sin 
1 



r 

sin ^ 



R + r sin r 

R2\ 1 



+ 



r VR + ^ sin 
R/, R 



R + r sin 



r^/ R + r sin 



y 



• cos^ OdO 
R + r sin ^ 



+ 2 



sin^ R2 



R2\ _d^ 
r^ / R» + r sin 



(3) 



+ 



sin OdO 
r 



+ 



+ 



'B.dO 



(4) 



+ 



Now 



- f4^^ ^^ + f 



RdO 

^2 



cosO 



2 



+ 



R^- 



= 0+^-f..(4a) 



A • f d^ r 

Again / ^ , . ^ = / 

^ y R + r sm (9 / 



__ dO^ 

K + 2 r sm . cos ^^ 



do 



sec^ n (^ ^ 



R + 2 r tan ^ . cos^ ^ 



R sec^ ^ + 2 r tan ^- 



seC 



/ 

dO ( put tan ^ = it, then 



R(l + tan2 j +2rtan2 



du = Jsec^ s 



542 



THE STRENGTH OF ^lATERIALS 

du 



2 
R 

2 
R 



du 
du 



J R (1 + «2) +%ru ^ J R + 2 r M + R m2 
J { ' 



d u 



u + ^) +(l 



R2 



2^ 
R 



■-(.'■-£)■ 



(5) 



This is of the form 



/d X 



■/ 



dO 



R ^ rsin^ R /: 



+ 13' P 



tan 



tan 



X — a 



U -r 



R 



\ ^ R2 

— B 

WTien ^ = '1. tan = 1 = w 



^ = — ^ tan;^ = — 1=2^ 

z, z 



1 - 



R2' 






cZ^ 



R + r sin ^ 



x/R2-r 



tan-^ 



1 



R 



r . \ 



I' r'- 
■M ^ ~ R2- 



tan 



R 



V^ ~ R2' 



/..n-_^L 



VR2 - r2 I VR2 - 



4- tan 



1 _AilJL.\ 

\/R2 - 7-2] 



(6) 



, _a /R +r , ^ , /R - rl 

The portion inside the bracket is of the form 
tan'^ y + tan"^ 

* See Lamb's Infinitesimal Oulculus (Camb. Uiiiv. Press), p. 172. 



CURVED BEAMS 543 

If tan a = -, cot a = ?/ = tan ( ^ — 

y ^ V2 



y 

.' . tan~^ ?/ = — — a = ^— tan"^ 

^2 2 y 

.' . tan~^ y + tan"^ = t^ 



(6) becomes / 



dO 



+ 



R + r sin ^ VR^ — ^2 

w 

2 



_ 5^^^ ^"^ n^ -r 



r^jR+rsine r^ VR^ — 

TT VR2 - 7-2 



r^ 



+ 2 



/.x J /^ X r Gos^ede Rtt TrVR^-r^ 

.'. from (4) and (4a) / --^-- ^ — ^ = -- „ --o 

^ ' ^ ^^ R + r sm (9 r^ r^ 



+ r 

m "^' ^ 



(2) R f^^ =27rR2-27rR VR^ - ^' 

^ J^ + y 

In (1) A, = 2 TT R2 - 2 TT R VR^ - r^ - ir r^ 
2 R2 2 R /7RV 



= .r^i-^---7(7T-l-l}----(^) 



I r^ r 



Correction Coefficients for Ordinary Beam Form- 
ulae.^ — -If the bending moment on an ordinary beam is M, 
we have for a rectangular section the bending stress 

, _ 6M 

' ~ hd'' 

and for a circular section of diameter d 

_ 32 M 



544 



THE STRENGTH OF MATERIALS 



Let /,, /o be the correct stresses at inside and outside 
respectively of the curve, then we may write 

/. = " / 















































































































2-0 














































































































































1-8 














































































































































re 


V 


. 


































\ 




































X 


^ 


C'l 


re 


le 




























\\ 
































inside. 




\ 


\ 
































\ 


\\ 




































V 


\ 




























•*3 








\^ 


s 


























<» 

1 I-a 

o 








/ 


<^ 


^ 






























/ 


































f 


?ec 


iro' 


pgi 


e 




^^ 




:::;::;;- 














o 


































— 


Con 





































Values of It ^ cZ 



2*5 



5*5 



4*5 



Fig. 255. — Correcting Coefficients at Inside. 



where a and ^ are correcting coefficients by which the 
stresses calculated in the ordinary manner should be multiplied 
to give the corrected values. 



CURVED BEAMS 546 

The following values of a and ^ have been calculated — - 



•75 
1 
2 
3 
4 
5 


Circular Section. 


Rectangular Section. 

1 


a 


•62 
•70 
•84 
•89 
•91 
•93 


a 


/3 


211 
1-62 
123 
114 
110 
1^08 


1-92 
1-52 
1-20 
112 

1-09 
107 


•65 
•73 

•85 
•90 
•92 

•95 j 



These figures are plotted in Figs. 255, 256. 
10 



'90 



o '80 

00. 



1 70 



p 

a 



•60 

































































































































_ 


- 


-;::: 




— 




















^^ 


-— - 




^^ 


■ — 






















_^ 


-^ 


--^ 


"^ 
















Re 


cYa 


na\ 


3, 


,^ 


--^ 


-^ 
























t 




^ 




-^ 


























V 


A 


y 






























} 


'/ 
































// 
































/ 


/ 
































// 


-^{ 


"ir 


ok 


i. 


























y 
































/^ 


/ 
































// 


































/ 










































































































































1 


1 


5 


2. 




Z 


5 


% 


J 


3 


•5 


^ 


1. 


4 


•5 


5 



Values of E. -^ tZ. 

Fig. 256. — Correcting Coefficients at Outside. 



The nearness of the curves for the two sections shows that 

the same coefficients may be used as a first approximation" 

for other sections. 

^ Andrews-Pearson Formula. — In this theory,* published 

* A Theory of the Stresses in Crane and Coupling Hooks, Draper's 
Company Research Memoirs, Technical Series 1 (Dulan & Co., Lon- 
don). For other experimental investigations see " Maximum Stresses 
in Crane Hooks," by Professor Goodman, Proc. Inst. C. E., Vol. CLXVII 
(1906-7); "An Investigation of Strength of Crane Hooks," American 
Machinist, Vol. 32, October 30, 1909. 



546 THE STRENGTH OF :\L\TERIALS 

in Um)4 by the author and Professor Karl Pearson, F.R.S., 
a correction is made for the fact that owing to lateral strain 
it is not quite correct to say that y = y^ 
The resulting formula is 






^r\ 



where 



A 71 = 



Ay, 



' R 



1 + 1 



R 



1 -^ ^ 
R 



i/\i-'j 



For simplification of resuJts we write 

73 = 7i - 72 
For a rectangular section we have 

_ R <Y 2R \v _ / 2 R .^> 

_ R )7 2R + t^y-^ /2R -d.^-^\ 
^^^ ~ (1 ->7)c/\^ 2R y '' 2R / ' 



Then A y, = ^ 



R 



A somewhat similar graphical construction can be employed 
to that in the Winkler formula, but still greater care has to 
be taken to ensure accuracy. 

These formulae are extremely troublesome to use and require 
the utmost care to avoid arithmetical error, and the additional 
accuracy over the Winkler formula is so small that it is 
doubtful if the revised formula is worth the additional trouble 
and risk of error. This point is dealt with very fully by 
Professor Morley in Engineering, September 11 and 25, 1914. 

The Strength of Rings and Chain Links. — In deter- 
mining the stresses in a ring or link we have, in addition to 



CURVED BEAMS 



547 



the problem of finding the stresses in a curved member, that 
of finding the bending moment at any point. 

Approximate Theory of Circular Ring. — We will 
calculate the manner in which the bending moment varies in 
a ring in which the effect of the curvature on such variation 
is negligible. By designing the ring for the bending moment 
thus obtained, allowance for the curvature effect upon the 
stress by the correction coefficients given in Figs. 255, 256, we 
shall obtain a very fair approximation to the stresses. 

Consider a radial section intersecting the centre line at c 





^^\ 






/ ^ x^'"'^ 


r-\ \ \ 


/ ' /^ 


WX \ \ 


I ' /^ 


\ ^ \ 


{'•■ I \ 


] • 1 


\ 1 \ 

\ ' \ 
V \ \ 


) '] 


\ ^ ^'"~" 




X N 


,^ / 




-^ 




Fig. 257. — Circular Rings. 

(Fig. 257). The stresses in each quadrant will be similar by 
symmetry. 

The bending of the link must be such that the sections 
o E, o B remain parallel to their original positions. 

But the angular change due to bending is given by the 

/M 
^^d s (see p. 360) 

/m d s 
J EI ^^ 



we have 



(1) 



Now the B.M. on the section 

o c = M, = B.M. at B - '' • ^ c 



w 

2 
Rsin^ 



(2) 



548 THE STRENGTH OF MATERIALS 



M„ - Y R sin ^ ] . R d! (9 



_.yM..^y-^M^ 



EI 



EI 



M„ . R ^ + ^ R2 cos ^ 



/MT.Y-_^^^WR^^_^^| ^ 1 



= -[M,R(;i-0)+ "o (o-i)y X 
M„R 



TT W R2\ 1 



2 2 ; EI 
If this - 0, M3 - ^^ - -318 W R (3) 

TT 

W R 

.• . M, = M, - -\p - W R (-318 - -5) 

CI 

= - -182 WR ..(4) 

The point of zero bending moment is given by putting 
M, =0 

.• . in equation (2) -318 W R - '5 W R sin ^ =. 

sin B = — p- = '636 
•5 

i.e,. 6 = 39" 5 degrees approximately. 

W sin ^ W 

The direct stress at b = ; at c = — . ; and at E = ^. 

The mean shear stress at B 

W ^ Wcos^ . ^^ „ 
= -^ ; at c = ^ . - and at L = 

2 ' 2A 

Numerical Example. — Find the safe load upon a mild steel 
ring of 4 inches mean diameter formed of round rod 1 inch in 
diameter. We will take the safe tensile stress in the material 
as 7 tons per sq. in. 

Let the safe load be W 

Then M, = 318 W R = 636 W 

. • . By ordinary bending formula 

"^~32^' = -636 W 



CURVED BEAMS 549 



In our case , = 2 
a 

.' . from the table on p. 545 a = 1-23 
.-. -636 W X 1-23 = ^^ 

•'• ^ = 32 X -636" X 1-23 = '^L^^^^^^J- 

Considering the section at E and taking this value of W 
we have 

Direct stress = ^^ ^ rrxn = '^^ *on per sq. in. approx. 

M, = -182 WR = -364 W 

, .. , 7 X M, 7 X -364 . ^^ 

bending stress = — j^— - = — ^636~~ ^ approx. 

.* . total stress = 4*00 + '56 = 4'56 tons per sq. in. 
Chain Links with Straight Sides. — We can apply as 
follows the same approximate theory to the determination 
of the stresses in a chain link composed of semi-circular ends 
and straight sides. Considering one quarter of the link as 
before, we have that the angular change due to bending 
between the points B and e (Fig. 258) must be zero. 

Between f and B we have as before 



Angular change = / ^j^^ 



'^{m ^_wr^\ 1 

im„. . - 'EI 



The bending moment will be constant over the straight 
part of the link. 

. * . Angular change between E and f = ^^ . ^ 

WR 

and M^ = M, = M« - ^- as before 



, /.. R^ WR2 M,L\ 
/. wehave^M..-^ 2^" + ^ ) 



Rtt WR2 . M.. L\ J^ 

EI 



,,^ Rtt WR2 M„L WRL\ 1 



2 2 ' 2 4 y • EI 



550 



THE STREXC4TH OF MATERIALS 



If this is zero, 



WR(R + 



M. = 



ttR -h L 
W R /2 R -f L 



2 Vtt R + L 



Fig. 258. — Oval Chain Links. 



_ WR f 2R ^ L _ 1 

~ 2 I TT R -^ L ^' 



WR/2R--R, WR^, 2-- 

""2 WR+L 



(5) 




WR^ 11416 \ 
2 VttR+L;" 2 VttR+L/ 2 WR+L/"^^ 

If L = these formulae reduce to the same result as in the 
previous case. 

Experiments on Chain Links. — The strength of chain 
links has been investigated ver}' thoroughh^ by Professors 
G. A. Goodenough and L. E. Moore,* who give a very complete 
theoretical treatment of the subject. 

* University of Illinois Bulletin, Xo. 18. 



CURVED BEAMS 551 

Their summary and conclusions are as follows — 

1. The experiments on steel rings confirm the theoretical 
analysis employed in the calculation of stresses. 

2. The experiments on various chain links confirm the 
analysis and show that the pressure may be taken as uniform 
over the arc of contact. 

3. A load on the link produces an average intensity of 

stress K -. in the cross section of the link containing the minor 
2 A ^ 

axis, and with an open link of usual proportions the maximum 

tensile stress is four times this value. 

4. The introduction of a stud in the link equalises the 
stresses throughout the link, reduces the maximum tensile 
stresses about 20 per cent, and reduces the excessive com- 
pression stress at the end of the link about 50 per cent. 

5. The stud-link chain of equal dimensions will, within the 
elastic limit, bear from 20 to 25 per cent, more load than the 
open-link chain, but the ultimate strength of the stud link 
is probably less than that of the open link. 

6. In the formulae for the safe loading of chains given by 
the leading authorities on machine design, the maximum 
stress to which the link is subjected seems to be under- 
estimated and the constants are such as to give stresses from 
30,000 to 40,000 lbs. per sq. in. for full load. 

7. The following formulae are applicable to chains of the 
usual form P = 0'4:d^ s for open links 

'P = 5 d^ s ,, stud „ 

where P = safe load, d = diameter of stock and s the maxi- 
mum permissible tensile stress. 



CHAPTER XX 

* ROTATING DRUMS, DISKS AND SHAFTS 

Thin Rotating Drum or Ring. — If a thin drum or ring 
of radius r (Fig. 259) rotates T^ith a velocity v, there will be 
acting on each unit length of the rhig a centripetal pressme 

p equal to , where iv is weight of unit length of the ring. 

1/ 

Thus pressure p will cause a hoop stress / and on any dia- 




FiG. 259.— Thin Rotating Cylinder. 

metral section the resulting force tending to cause bursting 
of the ring will be equal to p d, as in the case of thin pipes 
dealt with on p. 115. The force resisting bursting will be equal 
to / X 2 A w^here A is the cross-sectional area of the ring. 

We have, therefore, / x 2 A = p d = '— = 

, _ 2. w V- _ ic V 
''' f ^ YglK ~ gA ' 



ROTATING DRUMS, DISKS, AND SHAFTS 553 

Since w is the weight per unit length of the ring, we have, 
if p is the weight per unit vohime of the material, w ^ p K, 
since the volume of a unit length of the ring is 1 x A 

• f = P^ 
9 

If p is the weight in lbs. per cu. in. of the material, v is in 
feet per sec. and g = 32'2 feet per second per second, we have, 
bringing everything to inch units 

, p ?;2 X 12 X 12 12 p v2 

^ =^ ~3F2'^12" ^ "32^ ^^'- P^" '^- ^^• 

= f^approx. for cast iron. 
= qTk approx. for mild steel. 

This stress is often called the centrifugal stress. 

Numerical Example. — At what peripheral speed may a 
thin mild-steel ring be rotated if the centrifugal stress is not to 
exceed 16,000 lbs. per sq. in. ? 

We then have 

16,000 = ^ 

v^ = 9-5 X 16,000 
V = x/9*5 X 16,000 = 390 feet per sec. approx. 

Revolving Disk. — The consideration of the stresses in a 
revolving disk bears considerable resemblance to that of the 
stresses in a thick pipe, but it presents greater difficulty. If 
the disk is uniform in breadth and such breadth is compara- 
tively small we may proceed as follows — 

Considering, as in the case of Lame's theory (p. 510), an 
elemental ring at radius x of thickness 8 x (Fig. 260) and of 
unit breadth, we have a resultant centrifugal tension on the 
section given by 

-ij, w v^ J 2wv^ 
±,= ~ — . a = 

9/ 9 



554 THE STRENGTH OF MATERIALS 

If the angular velocity = o), i; = o) x 

2 9 

Jb,. = . 6 X since iv = p h x 

9 

For convenience we will write -- ^ a 

9 

.-. F. = 2qo>^x-Sx (1) 

Then by the same reasoning as in Lame's theor}^ we shall 
have 

Force tending to cause bursting of ring 

= F, + {p + Sp).2{x + S x) 
Force resisting bursting of ring ^2fSx + p.2x 




Fig. 260. 



These must be equal 



• ' • -^ + iv + ^P) (^ + S x') ^ j Sx + px 

neglecting products of small quantities 

F 

(V + p X + p S X + X ^ p = f S X + p X 

F 

. • . if — p)Sx=xSp+ 2 

= ^ + g -2 x^^ (by (1)) 



ROTATING DRUMS, DISKS, AND SHAFTS 555 

. • . In the limit f = p + ^/- + ^ co^ .t^ (3) 

Next consider the strains. If the radius x increases to 
{x + u), the circumference increases from 2 tt a; to 2 tt {x + u) 
. ' . increase in circumference = 2 tt u 

. • . Unital circumferential strain = ^ = - 

Z TT X X 

Also the thickness of the ring increases from 8a:;toSa: + Su 

o IJ u U 

.' . Unital radial strain = ^— = -, — in the limit. 

ox dx 

Now the principal stresses acting in an element of this ring 

are / and p 

. • . (as shown on p. 25) 

Unital circumferential strain ^ ^ [f — r] p) 
Unital radial strain = ^ {p — v f) 
i-e. -^ == ^ (/ -'V2?) (4) 

§-:=e(p-''/) (5) 

. • . solving these two simultaneous equations we have 

E fu . 7] du 



^-,-r^-)a" + fS <^) 

Putting these results in (3) we have 
E /u 7] du 

(1 - rj^) \X "^ ~d^ 

_ E /rj U du\ x'E / 7] U rj d u d^ u 

{I — r)^)\ X dx) \ — y]'^\ x^ X dx d x^ 

+ g w^ x"^ 

E 

i. e. multiplying through by t, ^ 

U , ridu 7] U , du 7]U , 7]d u xd^u {\ — rr-) „„ 

X dx X dx X dx dx^ E 



556 THE STRENGTH OF MATERIALS 

To solve this differential equation we write it 
x^ d^u , xdu (1—772) 

Now assume u = C x^ 

= X^.6CX+X.3CX^-CX^+ ^^ ^^ . g 0)2 x3 

Equation (9) will be solved if C = - ^^~~f2 ^ "^^ .... (10) 

This equation is of the"! kind dealt with in Forsyth's 
Differential Equations (Macmillan), §§ 38, 39. 
The '' complementary function " is 

x^ d^ u X du _ 

d x^ d X 

d^u 1 du '^ _ ci 
' ' d x^ X d X x^ 

d^u d /u\ _ 
' ' d x"^ d X \x/ 

Integrating, we have 

-J 1- - = constant = Cj (11) 

ax X 

To integrate again, write it 

X a u ^-j 

_ + w =r (Ji a: 
dx ^ 

1. e. --^ — - = Ci .T 
dx 

C a;2 
. • . integrating u x = ■— \- C2 

■ , •••I = ^' + & (12) 

. • . putting this in (11) 

4j* _ Ci _ C2 ,,0 

dx~ 2 x^ ^ 



KOTATING DRUMS, DISKS, AND SHAFTS 557 
We also have from " particular integral '' u = C x^ 

- = C X2 

X 

t^ = 3 C a;2 
a X 

. ' . adding the complementary and particular integral we 
have 

^ = C a;2 + ^^ + % 
X 2 x^ 

(1 - v^) q 0) 2 ^^ , Ci C2 ,..s 

= 8E + T + ^ ^^^^ 

du _ 3(1— 7)^) q 0)2 x^ , Ci _ C2 /, K\ 

.dx~ 8E "^2 0:2 ^ ^ 

Special Cases. — (1) External radius R, internal radius r. 
We must have 2? = at the inside and outside 

. • . p = for a; — R and a; = r 

. * . at inside in equation (7) 

o — -^ /rju du 
~ 1 — rj^\ r dr 
E r (1 -r}^)qo,^r^ , C^rj 0^7} 



r 



'•^' ^ - (1 - ^2) \ ^g-E -^ + 2 + 

_ 3(l-ry2)go,2^2 Ci _ C2\ 

8E "^2 r2j 

i.e.5(l+>;)-jMl-.)=^^-^^^^^3+,) ..(16) 
Similarly at outside where a; = R we shall have 



^ (1 - 77) (r2 - R2) . 8 E 

(1 + r)){3 +r;)ga)2R2r2 



8E 
Putting this in (16) and simplifying 



(17) 



^^_ (1 - r,) {S + r,) q o>^ (R^ + r^) ^jgj 

4 E 



558 THE STRENGTH OF MATERIALS 

Now put these values into the equations (6), (14) and (15) 
for / 

- (r=:^2) 1^ - 8 E + 2 (^ + ^^ + x^ ^^ - ^^ 

3 77 (1 — rf) q la^ X^\ 

8E / 

+ (3 + ';) -/- - 3 , x^} 

Similarly to obtain p we use equations (7), (14) and (15) 
and take the given values of the constants. 
E ^riu dn\ 

^ {I -yj^)\x ^ dxf 

-{l-r,')\- 8 E + 2 (1 + ^) - :.2 (1 - V) 

_ 3^1 - 7y2) qj^^ 
" 8E f 



R2r 



(3+^)- 2 -3 a: 



X 



p w2 (3 + 7?) f-^ R^r^ ^ 



It is clear from equation (19) that the greatest hoop tension 
occurs at the inside where x — r 

9 
and if r is very small 



••• /-.v. = ^ " {(3 +v)^' + a-l)r'} (21) 



_ p w2 R2 (3 + ^) 
/max. - -4^ " — l^^J 



ROTATING DRUMS, DISKS, AND SHAFTS 559 

(2) Disk without central hole. — If the disk is a solid one of 
radius R, ^ = for a; = R and u the circumferential strain 
must be for x = 

.'. from equation (14) Cg = 

.'. Equations (14) and (15) become 

u _ (1 - rf)qio^x''- Ci 

x~ ~ 8E +2 ^^"^^ 

du 3 (1 — 7)^) q iJi^ x^ , C, ,^.- 

di=-- 8E- + 2 (24) 

. • . Since p = for x = H, equation (7) becomes 

E f7?Ci 7y(l -7?^)go>2R^ Ci 3{l -yj^)qu>^U^ '\ 
(1_^2)\ 2 "SE" ' "^2 8E j 

t. e. J (1 + ^) = ^^^^^ g co^ R2 (3 + ,7) 

(1 -r;)(3+>?)go>^R^ ,^.. 

^1 - ~ 4E ^ ^ 

. • .^Equation (6) becomes 

^ f _ (1 - rjf q oi^X^ (1 -^) (3 + r]) g (o^ R2 

'''{i-7]^)\ 8E "^ 8E 

_ 3 ry (1 - 7/2) g (o^ a; ^ (1 - ry) (3 + 7?) g (o^ R^l 

8E ' "'"'^ "8E ^ j 

= ^f 1(3 + .7) R2 - (1 + 3 r;) a;-^} 

= ^ 1(3 + >?) R^ - (1 + 3 r;) a:2j (26) 

Similarly equation (7) becomes 

P = ''^g {B + v) Cii' - ^') (26) 

Each of these is a maximum at the centre where x ^ 0, 
where we have 

Lax. = Z^Huax. = Sj- (3 + ^) (27) 

This is exactly one half the value given by equation (22) 
for a disk with a very small hole, so that on this theory a disk 
has its strength reduced by one half by having a hole made 



560 THE STRENGTH OF ]MATERL\LS 

through it. For this reason the De Laval turbine drums are 
made without a central hole. 

If V is the peripheral speed at the outside we have r = w R 

.-. Equation (27) gives / = ^- (3 -f v) ^ti^h is ^^-—^ of 

the stress in a thin drum of the same external radius (of. 
p. 553). 

Taking , = 1, i^+2) = g 

NuMEEiCAL Example. — At wlmt peripheral speed may a 
narrow mild-steel disk he rotated so th<it the maximum tensile 
stress sh-all not exceed 16;000 Ihs. per sq. in. (a) // it h^s a small 
hole in th-e centre, (6) if it is quite solid ? 

(a) By equation (22) 

16,000 = ^'. (3-25) 
4:g 

~ 9-5 ^ 4 

., 9-5 X 16,000 X 4 
}'- ^=. 

3-25 



9-5 X 16,000 X 4 ,o^^ . 

^7^.; = 432 feet per sec. approx. 

(6) by equation (27) 

16,000 = ^J' . 3-25 

.'. r- = 432 V2 = 612 feet per sec. approx. 

Rotating Disk of Uniform Strength. — A problem which 
is of interest in turbine design is that of finding the shape of 
a disk which will have the same stress throughout. ^Ye have 
seen already that the maximum stress occiu-s at the centre so 
that the disk ought to be broader at the centre than at the 
edge. 

Referring to Fig. 261 and considering the elemental ring of 
breadth b at radius x increasing to (6 -r S b) at radius {x + h x) 
we have the principal stresses each equal to / which does not 
vary radiaUy. 



ROTATING DRUMS, DISKS, AND SHAFTS 561 

Force tending to burst ring = F, + 2 / {x + Sx) {b + Sb) 

. • . If these are equal, neglecting products of small quantities 
we have 

¥,. + 2fxb+2fbSx-{-2fxSb=2fbx + 2fbSx 

¥, + 2fx8b = 
Sb^ 



wv^ 



NowFe = ^^ .2x = 
gx 



j,{h+'-iY 



a; . 0)2 x^ 



2 ph hx . (n^ x^ 
9 




Fig. 261. — Disks of Uniform Strength. 



. * . Dividing hj 2 x f 

pb (M^ 



fg 

in the limit 



.x^x -\- hb = 

pb 0)^ , db r. 
'^^ , ^ _j- = 
J g ax 



p w^ 



JC2 



The solution of this is 6 = C e a' 2~f 

At the centre where x = 
b ^C = b. 

.\b = boe g' 2}' 



oo 



562 



THE STRENGTH OF IVIATERIALS 



The thickness at the outside is given by 



-P .'^^ .H2 



If therefore b^ is given we can calculate b., by 



p . u>-^ 



bj^e' ^^/ 



R-i 



/ 

. • . At any radius ^ =b^e 



£7-2/ 



Whirling of Rotating Shafts; Critical Speeds.— If 
a shaft rotates at a high speed, the lack of mathematically 
exact balancing results in an eccentricity of load which causes 




Fig. 262.— Whirling of Centrally Loaded Shaft. 

centrifugal forces to be induced and these centrifugal forces 
will cause deflections which increase the eccentricity to be 
increased ; this increased eccentricity causes further deflection 
and so on, the deflection increasing indefinitely and giving rise 
to whirling at certain speeds called critical speeds. 

In certain cases the whirling speed is the same as the natural 
frequency of transverse vibration of the shaft. 

The centrifugal forces may be regarded as having a neutralis- 
ing effect upon the elastic forces tending to return the shaft 
to its natural shape, so that when whirhng occurs the effective 
stiffness of the shaft is reduced to zero. 

Flexible Shaft loaded at Centre. — Referring to Fig. 262 
let a shaft a b of length I be loaded at the centre with a disk 
of weight W and let the shaft be provided with flexible bearings 



ROTATING DKUMS, DISKS, AND SHAFTS 563 

which do not interfere with the natural deflection of the shaft. 
Then if 8 is the deflection caused by the centrifugal force and 
e is the eccentricity of the load, i.e. the distance from the 
centre of the shaft to the centre of gravity of the load, the 
effective eccentricity is (S + e). If the angular velocity is w 

. • . Centrifugal force = F = — ^ 

9 
Ai . F^' ■ T? 48EI8 
^^^^=48EI '•'■^- ' p ' 

48EI8 Wco^, , , 
/48EI W(o2^ Wco^e 



l^ 9 y 9 

W 0)2 e 48 E I 



.8 = 



9 ' l^ 9 



4:SElg -w^WP ^^^ 

From this equation it is clear that 8 will become indefinitely 
great if 48 E I (/ - w^ W Z^ = 

^.6. if.= Z^^LX (2) 

This value of w gives the critical speed. 

For mild steel E = 30 x 10^ lbs. per sq. in. and g = 
32*2 X 12 ins. per sec. per sec. ; if therefore W is in lbs. and 
I and I in inch units 



48 X 30 X 106 X 32-2 x 12 . 1 



WP 

= 746,000. / radians per sec (3) 

74,600 X 60 /nr 



2 TT MWP 

= •711 X W^L revolutions per minute (4) 

For a round shaft of diameter d inches we have I 



ird^ 



64 

•158 X 10^ d-^ 

n = j-^=^^ — 



564 THE STRENGTH OF :VIATERIALS 

Working from the transverse vibration we have, as on p. 337, 

^ = 2 77 / ' Weight 

^ g X force to cause unit displacement 

48 E I 



. • . Force = W 
.-. t = 2- 



V48EI^ 



for unit deflection 



Frequency 



1 1 

't 



= o-a/—^-^ per second 

2-V WZ3 ^ 

-^n J^^^V^T minute. 
77 X 60 \ WZ^ 




Fig. 263. 

This,, for the given value of E, is exactly the same result as 
is obtained in equation (4) above. 

If the critical speed is exceeded either by providing guides 
which prevent the excessive deflection or by speedmg up so 
quickly that the inertia of the shaft prevents the dangerous 
deflections from developing, the shaft will '" settle down " 
and nui smoothly in a deflected form (Fig. 263), the weight 
rotating about an axis which gradually approaches its centre 
of gravity as the speed increases. This fact is made use of in 
the flexible shaft of the De Laval turbine. 

If t-j is the critical velocity we may put in equation (1) 



ROTATING DRUMS, DISKS, AND SHAFTS 565 



i. e. 8 = 



W P 0..2 -WP 0.2 o.;2 



(u2 — O) 2 



•e (5) 



This gets numerically less as o) increases, so that as the 

speed increases more and more the shaft tends to straighten out. 

If the shaft is horizontal and the weight is perfectly balanced, 

W P 
and there is an initial deflection S„ = j^ ^ y , this will give rise 

to a centrifugal force which causes an additional deflection 8^ 

...r = Wo>2 8.) 
g 

FP 
^""^ ^1^48 EI 




Fig. 264.— Whirling of Unloaded Shaft. 

48EI8i Wo)2 . , W(o2 . 

P 9 g 

'48 EI Wo)^ _ Wco^ 
9 ^ ~ 9 



, /48EI Wo)^ W(o2 , 
I.e. Oj — ^ = . do 



Woj2 ^ 48 EI Wo.2 

9 P 9 

W 0)2 Z3 8, 



~ 48 E I ^ - 0.2 w ^3 (6) 

This, as one would expect, gives the same result as before 
"with e = 8„. 

Unloaded Shaft. — In this case we obtain at certain critical 
speeds a condition of insta?jility which is very similar to that 
which occurs in a loaded column. 

Suppose that the shaft of length I is initially straight and 
that due to some cause it becomes deflected so that at some 
point p (Fig. 264) at distance x from a convenient origin, say 
the centre point c, the deflection is y. 



566 THE STRENGTH OF MATERIALS 

If w is the weight per unit length of the shaft this deflection 

will cause at the point p a load equal to per unit length. 

But since the B.M. diagram is the second integral of the 
load diagram (p. 149) we shall have 
dyM. _ ww^y 
dx^ g 

Moreover, .pr^ = ^-^ 
EI d x^ 

w ui^ y d^ y 
''' EI> ^ dx^ 

2 

putting -^ J- == m*j we get the differential equation 

»'2'=rf.^ w 

The general solution of this is 

2/ = A cosh m X + B sinh mx -\- C cos m x -{- D sin m x . . (2) 

Ends freely supported. — If the ends are freely supported, the 
deflection and B.M. are each zero at each end. 

. • . 2/ = when x = — ^ and x = + ^ 

-,%, = when .r = — ^ and x = ^ 
dx^ Z ^ 

d v 
also the slope -^ = when .r = 

Cv X 

7nl ^ . , ml ^ ml ^ . ml 

.-. = A cosh -^ + Bsmh- 2" +C cos - -2- + D sm — ^ 

^ ml ^ . ^ rnl , ri ml -^^ . ml 
= A cosh -o^ + B smh ^^ + C cos ^ + D sm ^ 

^ ml 1 ml ^ ^ ^^c "^ ^ 

Now cosh Y = cosh — ^ ; cos ^ = cos — ^ 

ml . , ml . ml ^\r.'^^ 

smh - 2 = - ^1^^ 2^ ' ^^^ "" 2 ^ ~ sm 2 

D a nd B each = 
. • . A cosh '2 + C cos 2 =0 (^) 



ROTATING DRUMS, DISKS, AND SHAFTS 567 

. • . our equation becomes 

y = A cosh m X -\- C cos m x 

,-r d cosh X . , d cos x 

Now — ^ = smh X ; —^ = — sni x 

Cb X Cv X 

d sinh X , d sin x 

— -^ = cosn X ; — ^ = cos x 

Cb X Cv X 

CJ tJ 

. ' . -7-^ = Am^ cosh m X — Cm^ cos m x 
dx^ 

.-.putting X = + ^OT - - 

.' .0 = Am^ cosh ;-- — C m^ cos — 

2 2 

i. e. A cosh -^ — C cos -— = (4) 

2 2 

.• . Comparing this with (3) we see that A must = 

.' .y = C cos m x 

Further, C cos -~- = 



ml TV , 
— = 2 rtc. 



Taking the lowest vahie 



l^"" 



m (w 0)2 YJ 



EIi7 

o) = ^ A / ^ radians per second 

l^ \ w 

. • . If ?i is the number of revohitions per minute 

Itt n IT n • 30 (D 

30 TT /EI a 1 • 

• ' • ^ 72 " A/ ^^ revolutions per mmute (5) 

If the shaft is of diameter d inches and I is in inches 
Taking E =- 30 x 10^ lbs. per sq. in. 

g = 322 X 12 ins. per sec. per sec. 



568 THE STRENGTH OF IVL.\TERIALS 

64 

tc = -f_ X -28 lb. 
4 

, 4-8 X 106 f/ ,^, 

we have n = (6) 

9 — "^ — 4. — 

The higher critical speeds -v^ill occur for m = *" " , " , " , 

^ JU ^ 

etc., giving values of n multiplied by 4, 9, 16, etc. 

Both ends fixed-. — In this case ,^ = for x — + in 

dx ~ 2 

addition to the condition that v = for x = 

2 

.• . B and D are each equal to as before 

and A cosh ^^ -f- C cos -^ = 
2 2 



when .r = - , — " = m A suih ~-^ — m C sin - = 
2' dx 2 2 



I d y . . ^ ml ^ . 7?i Z 

^ — ni A sum — — ^n n oi^-. 

2 

«. e. A sum ^ — C sm -- = 

. m I 



' ' C . , m I 
suih — 

m Z 

A ""^ 2 

Also from (3) 7^ = — — , 

cosh 

, ml , -, ml 
.• . — tan = tann 

iu ^ 

3 TT 

The solution of this gives w Z = 474 = " aj^prox. 

Taking the approximate value, this will give the first critical 
speed about nine times that in the case of the freely supported 
shaft. 

Dunkerley's Empirical Formulae. — Professor Dunker- 
ley,* who was one of the first investigators of the theoretical 
* Phil. Trans. Roy. Soc. 1895, Liverpool Engineering Society, 1894-5. 



ROTATING DRUMS, DISKS, AND SHAFTS 569 

and actual whirling speeds of shafts, has given the following 
empirical formulae which agreed very well with his experiments. 
Let oji be the critical angular velocity for a given unloaded 
shaft; let o)^ be the critical angular velocity of the same 
shaft carrying a wheel at any position, neglecting the mass of 
the shaft. Then the critical velocity w^ of the loaded shaft 
will be given by 



or if Uo, n^, n^ are the corresponding number of revolutions 
per minute 



Tio^ n-^^ n^ 



If a second load be keyed at another position, the critical 
angular velocity of which is wg with the first load removed 
neglecting the weight of the shaft, then 



(Oi (Oo 0)r 



or m general — = ?, — 



For further information on this subject, the reader may 
refer to Professor Dunkerley's paper and to Stodola's Steam 
Turbines (Constable) . 



EXERCISES 



CHAPTER I 



1. A tie rod in a roof structure has to stand a total pull of 40 tons. If 
the stress in the material is to be not greater than 5 tons per sq. in., find 
a suitable diameter. Ans. 3| ins. diam. 

2. Taking the shearing strength of mild steel to be 20 tons per sq. in., 
calculate the force necessary to punch a f in. hole in a f in. plate. Find 
also the stress in the punch. Ans. 29*4 tons ; 66' 7 tons per sq. in. 

3. A bar of mild steel fin. diam. and 10 ins. long stretches '00816 in. when 
carrying a load of 5 tons. Calculate Young's modulus (E) in lbs. per sq. in. 

Ans. 30 X 10^ lbs. per sq. in. 

4. If E is 29,000,000 lbs. per sq. in. for wrought iron, what decrease in 
length of a column 20 ft. high and 12 sq. ins. sectional area takes place 
when carrying a load of 36 tons ? Ans. '0556 in. 

5. What load in lbs. is hung on an iron wire 50 ft. long and *! in. diameter 
to make it stretch -^o- in. ? Ans. "076 lb. 

6. Plot a stress-strain diagram for the following test of a specimen from 
a mild-steel boiler plate — 



Load lbs 


4,000 
•0009 


8,000 
•0020 " 


12,000 
•0033 


16,000 20,000 24,000 


28,000 


Extension ins. . . 


•0044 •OOSe ^0070 

1 


•0082 



Load lbs 


30,000 


34,000 


36,000 


1 
40,000 44,000 48,000 


52,000 


Extension ins. . . 


•0103 


•016 


•7 ■ 


•19 ^30 -47 


•75 


Load lbs 


56,000 
1-3 


59,780 


54,900 


g^ , /Luads-1" = 10,000 lbs. 
ocaies (^Extensions— up to yield 


Extension ins. . . 


25 


2^9 


point 500 times full size. 
Beyond = 4 times do. 



Orig. dimens. Length = 10 ins., width = 1*753 ins., thickness = '64 in. 
Final „ „ = 12-9 ins. „ = 1*472 ins. „ = '482 in. 

Find stress at elastic limit, maximum stress. Young's modulus, and per- 
centage extension and reduction of area. 

57] 



572 EXERCISES 

7. In a plate girder the maximum intensity of stress at right angles 
to the vertical cross section of the web is 5 tons per sq. in., and the in- 
tensity of shearing stress is 2 tons per sq. in. Find the position of 
the planes of principal stress at that point and their intensities. 
(A.M.I.C.E.) 

Ans. 19° 20' aTid 70° 40' to vertical ; 5" 7 and 0*7 tons per sq. in. 

8. The limit of elasticity of a W.I. bar was found to be 20,000 lbs. per 
sq. in., the strain at that point being 0*0006; what was the resilience of 
the material? (A.M.I.C.E.) Ans. 6 in. Ihs. 

9. Two rods, one of copper and the other of steel, are fixed at their 
top ends, 24 ins. from one another, and hang vertically downwards. They 
are connected at their bottom ends by a horizontal cross-bar, and on this 
bar is to be placed a weight of 2000 lbs. If each rod is 18 ins. long, and 
if the diameter of the copper rod is 1 in. and of the steel rod f in., find 
where the weight must be placed so that the cross-bar may remain hori- 
zontal. E for copper = 16 X 10^ lbs. per sq. in. ; for steel = 29 X 10^ 
lbs. per sq. in. (B.Sc. Lond.) Ans. 11*9 ins. from the steel rod. 

10. A load of 560 lbs. falls through ^ in. on to a stop at the lower end of 
a vertical bar 10 ft. long and 1 sq. in. in section. If E = 13,000 tons 
per sq. in., find the stresses produced in the bar. 

Ans. 5*45 tons per sq. in. 

11. A bar of iron is at the same time under a direct pull of 5000 lbs. per 
sq. in., and a shearing stress of 3,500 lbs. per sq. in. What will be the 
resultant tensile stress in the material ? Ans. 6,800 lbs. per sq. in. 

12. In Question 11, find the resultant tensile stress from the strain 
consideration. Ans. 7,250 lbs. per sq. in. 

13. Find whether, in the problem of Questions 11 and 12, on the as- 
sumption that the shear strength of the material is 4 of the tensile strength, 
the resultant shear stress is more serious than the resultant tensile stress 
or strain. 

Ans. Res. shear stress = 4,300 lbs. per sq. in. Not so serious. 

14. Steel rails are welded together and are unstressed at a temperature 
of 60° F. They are prevented from buckling and cannot expand or con- 
tract. Find the stresses when the temperature is : (1) 20° F., (2) 120° F., 
taking steel as expanding '0012 of its length for a temperature change of 
180° F. E = 30 X 106 lbs. per sq. in. If the elastic limit is 40,000° F., 
at what temperature ould it be reached ? (A.M.I.C.E.) 

Ans. 8000, 12,000 lbs. per sq. in. ; 260° F. 

15. If the stress p at a point on one plane is inclined at an angle of 
60° to that plane and on a plane at right angles to the former the stress is 
a simple shear, find the principal stresses at the point and their direction. 

-4n5. 1 (l ± \/|) ; tan2e= y 



EXERCISES 573 



CHAPTER III 



1. In a roof truss a certain tie lias in it a pull of 3*05 tons due to the 
dead weight alone. When the wind is on the left of the truss it alone 
causes a pull of 5*5 tons in the same tie, and when it is on the right side 
it causes a compression of 1*2 tons. Work out what you would consider 
a satisfactory section for the tie if it is made of mild steel. 

Ans. 3 ins. x f in. flat. 

2. Estimate the dead load equivalent to a tensile dead load of 15 tons 
and a live load of 20 tons ; if the strain is not to exceed '001, find the area 
of section required, E being 13,500 tons per sq. in. 

Ans. 55 tons ; 4" 07 sq. ins. 

3. A 3-girder bridge to carry a double line of rails has an effective span 
of 38 ft. 6 ins. Find a suitable working stress assuming that the weight 
of the girders is 5^^ of the weight to be carried ; that the flooring weighs 
7 cwt. per ft. run of the whole width of the bridge ; that the permanent 
way, etc., weighs 160 lbs. per foot run for each line of rails; and the live 
load is 40 cwt. per foot run per line of rail. Ans. 5 tons per sq. in. 

4. What load, suddenly applied, will produce in a mild steel bar an 
extension of -^ of an inch ? The bar is 5 ft. long and 1| sq. ins. in section. 
Take E = 13,000 tons per sq. in. Ans. 4*06 tons. 



CHAPTER IV 

1. Two lengths of a flat steel tie bar, which has to carry a load of 50 tons, 
are connected together by a double butt joint. The thickness of the plate 
is I in. Find the diameter and the number of rivets required, and the 
necessary width of the bar for both chain and zigzag riveting. What is 
the efficiency of each and the working bearing pressure ? Make a dimen- 
sioned sketch of the joint. 

Ans. I in. rivets, 12 and 10| in^. wide ; 75 per cent, and 87 per cent. ; 
9*2 tons per sq. in. 

2. A diagonal tie in a lattice girder has to carry a load of 15|^ tons and 
is I in. thick. Using f in. rivets, find the necessary width of tie and 
calculate the number of rivets required (in single shear) and sketch the 
arrangement. Aiis. 5 J ins. wide, 7 rivets. 

3. Plates 1 in. thick are connected by a treble riveted butt joint, the 
pitch in outside rows being twice that in the others, and d = lin. Taking 
shear resistance in double shear = 1*75 times that in single shear, determine 
p for equal shear and tearing resistance. Find also the efficiency. 

Ans. 6 1 i7is. ; 85 per cent. 

4. For equal strengths in tension and shear calculate the pitch for a 



574 EXERCISES 

butt joint, given the following data: Plates 1 in. thick; rivets 1| ins. 
diam. ; two rows of rivets on each side of joint ; f, = 54,000 ; /< = 65,000 
lbs. per sq. in. Ans. 5g- ins. 

5. A steel boiler 4 ft. in diameter, and subject to a pressure of 200 lbs. 
per sq. in., is \ in. thick. Find the intensity of tlie circumferential and 
longitudinal stresses, the efficiency of the joints being 75 per cent. 

Ans. 12,800; 6,400 Ihs. per sq. in. 

6. Find a suitable thickness of plate and design a double riveted lap 
joint (longitudinal) for a cylindrical drum 5 ft. in diameter, subjected to 
an internal gauge pressure of 250 lbs. per sq. in. Take a working stress 
of 5 tons per sq. in. (A.M.I.C.E.) 

Ans. Plates 1 in. thick ; rivets 1^- ins. diameter ; 3 J ins. pitch. 

7. Calculate the thickness of shell of a boiler 4 ft. 6 ins. in diameter to 
resist a pressure of 150 lbs. per sq. in. Assume an efficiency of riveted 
joints of 70 per cent, and take the working stress as 6 tons per sq. in. 
(A.M.I.C.E.) Ans. -^^ in. 

8. Determine the stresses across the longitudinal and transverse sections 
of the plates of a boiler drum 3 ft. in diameter and J in. thick, subject to a 
steam pressure of 200 lbs. per sq. in., assuming that the drum is long and 
that it has no longitudinal seam. Ans. 7,200 ; 3,600 lbs. per sq. in. 



CHAPTER V 

1. A cantilever whose weight may be neglected carries isolated loads 
of 2 tons and ^ ton at distances of 5 ft. and 8 ft. respectively from its built- 
in end, the cantilever being 10 ft. long. Sketch shear and B.M. diagrams. 

Ans. Max. B.M. = 14 ft. tons ; shear = 2|- tons. 

2. A certain joist used as a cantilever weighs 18 lbs. per foot, and the 
max. B.M. which it can carry is 63 "56 in. tons. Find how long the span 
may be for the cantilever to be able to safely sustain its own weight. 

Ans. 36-3 ft. 

3. A beam of 12 ft. span carries loads of 3 and 4 tons at distances of 
5 and 8 ft. from the left-hand support. Draw the shear and B.M. curves. 

Ans. Max. B.M. = 15*66 ft. tons ; reaction, 3"91 and 3*09 tons. 

4. A beam of 25 ft. span carries a load of ^ ton per foot run, and an 
isolated load of 6 tons at a distance of 4 ft. from the left-hand support. 
Find the maxihium bending moment, and sketch the shear and B.M. 
curves. Ans. Max B.M. = 25*2 ft. tons. 

5. A beam of 40 ft. span carries a uniformly distributed load of 20 tons ; 
at points 11 ft. 3 ins. from each end isolated loads of 11 tons are carried, 
and between these points and each end additional loads of 4*5 tons are 
uniforml}" distributed. Draw the B.M. diagram. 

Ans. Max. B.M. = 250 ft. tons nearly. 



EXERCISES 575 

6. A beam 25 ft. long is anchored down at one end and rests over a 
support 6 ft. from the other end. It carries a load of 15 tons at the free 
end, and a uniform load of 5 cwt. per foot run. Sketch the shear and 
bending moment curves. Ans. Max. B.M.— 94*5 ft. tons. 

7. A beam 34 ft. long overhangs one support by 6 ft. and carries a load 
of 10 tons uniformly distributed. In addition it carries a load of 3 tons 
at the overhanging end and a load of 12 tons uniformly distributed along 
a length of 12 ft. commencing from the other end. Find the maximum 
bending moment on the beam. Ans. 62*2 ft. tons. 

8. A beam is laid horizontally upon two supports which are 12 ft. apart, 
and projects at each end 6 ft. beyond the support. A load of 2 tons is 
carried upon each of the projecting ends, and 1 ton at the centre of the 
span. What is the B.M. at the centre and at each support ? Sketch the 
B.M. diagram. (A.M.I.C.E.) Ans. 9 ft. tons ; 12 ft. tons. 

9. A plate girder is built of depth = J^ span. The maximum per- 
missible B.M. in ft. tons in such girder is roughly given by formula : 
B.M. = 7 X area of flange in inches X depth in feet. Find the maximum 
span for such a girder to carry its own weight : {a) neglecting its web 
altogether; (6) taking its web as half the sectional area of one flange. 
Neglect all angles, rivets, and stiffeners. Take steel as weighing 490 lbs. 
per cub. ft. Ans. {a) 1,536 ft. ; (b) 1,229 ft. 

10. A beam of 20 ft. span carries a uniform load of 5 cwt. per ft. run and 
an additional load of 9 tons spread over 12 ft. starting from the right-hand 
end. Draw the B.M. and shear diagrams. 

Ans. Max. B.M. 38*7 ; Reactions 8*8 and 5-2 tons. 



CHAPTER VI 

1. Find the moment of inertia about the centroid of an I beam 8 ins. 
deep, the width of flanges being 5 ins. The flanges are '575 in. and the 
w^eb '35 in. thick. Ans. 89' 1 in. units. 

2. A stanchion section consists of two standard channels 11 ins. X S^ ins. 
placed back to back at 6^ ins. apart and two 14 ins. X ^ in. plates riveted 
to each flange. Find the least radius of gyration. Ans. 4*12 ins. 

3. Find the radius of gyration of a hollow cylindrical column with an 
external diameter of 12 ins. and a thickness of 1 in. ; also of a solid square 
column 4 ins. hj 4 ins. Ans. 3*90 ins. ; 1*15 ins. 

4. A cast-iron girder has an upper flange 4 ins. by 1 in. ; a lower flange 
8 ins. by 1|- ins. and a web 6 ins. by 1 in. Find its moment of inertia and 
radius of gyration about an axis through the centroid parallel to the 
flanges. Ans. 195 ins.'^ ; 2*98 ins. 

5. A channel section has a base of 10 ins. ; sides 3 ins. ; the thickness of 



576 EXERCISES 

metal being f in. Find the position of the centroid and the moment of 
inertia about a line through the centroid parallel to the base. 

Ans. '726 in. from hase ; 6*62 ins.^. 

6. A column is built up of two I beams 10 ins. deep and with flanges 
5 ins. wide, the centres of the beams being 10 ins. apart. The area of each 
is 8*82 sq. ins., and the greatest and least moments of inertia are 145"7 and 
9" 78 in. units respectively. Riveted at the top of each pair is a plate 
12 ins. wide. Xeglecting the rivets, find the thickness of the plate if the 
greatest and least moments of inertia are the same. Ans. yV in. 

7. A column is built up of two channel sections 12 x 82^ X i^ in., with 
a plate ^ in. thick riveted to the flanges at top and bottom. Find the 
distance x apart that the channels must be for the moments of inertia to 
be equal about the two axes of symmetry, the width of the plates being 
ic + 7J ins. Ans. 9f iiis. 



CHAPTER VII 

1. A 20 in. X 7J in. joist is supported at both ends. The weight per 
foot of this section is 89 lbs., and the moment of inertia = 1,646 ins.*. 
Find the distributed load in a 25 ft. span which vnll cause a max. flange 
stress of 7 tons per sq. in. Ans. 29*7 tons net. 

2. The moment of inertia of a 12 in. X 5 in. X 32 lb. joist is 221 ins.*. 
Two such joists are placed side by side, and support a water-tank which 
weighs 1 ton when empty. Eflective span = 15 ft. What is the weight 
of the water in the tank when the stress in the extreme fibres of the joist 
is 6*5 tons per sq. in. ? Ans. 19*8 tons. 

3. Two 6 X 3 X ^ in. "fs are used back to back as a girder on which a 
light crane runs. Compare the safe load which such a beam would carry 
with that of a joist of same span, depth, width, and thickness of metal. 

Ans. Joist 5*36 times as good. 

4. Find the bending moment which may be resisted by a cast-iron pipe 
6 ins. external and 4^ ins. internal diameter when the greatest intensity 
of stress due to bending is 1,500 lbs. per sq. in. Ans. 21,750 in. lbs. 

5. A rolled-steel joist 16 ins. deep, with flanges 6 ins. wide and 1 in. thick 
(the web being | in. thick), is used to support a uniformly distributed load 
of 2 tons per ft. run. If the span is 12 ft. 6 ins., what is the maximum 
stress in the lower flange ? (A.M.I.C.E.) Ans. 4^'^ tons jjer sq. in. 

6. Find what diameter of axle should be employed if the wheels are 
4 feet 9 ins. apart and the loads on them are 7 and 3 tons respectively, the 
axle boxes projecting 9 ins. beyond the wheels. Draw the B.M. diagram. 

Ans. Max. B.M. = 58' 68 in. tons ; diameter = 5 ins. 

7. A gallery is carried by two 9 in. x 3 in. timber cantilevers, each 5 ft. 
long. What distributed load may the gallery carry if the safe stress is 
10 cwt. per sq. in, Ans. 27 cwt. 



EXERCISES 577 

8. Either of the following sections is available for a beam which is 
required to be as strong as possible : (a) Circular, 2 ins. diam. ; (6) rect- 
angular, 2 ins. deep, 1"178 ins. wide. Which would you use ? (A.M.I.C.E.) 

Ans. Circular. 

9. A cast-iron beam section is 20 ins. deep ; top flange 4 ins. x 1 in. ; 
bottom flange, 16 ins. X 1| ins. ; web, 1 in. Find the safe distributed 
load which a cast-iron girder of the above section, and of 20 ft. span, 
could safely carry. Take the safe stresses as 1 ton/in.^ in tension, and 
4 tons/in.^ in compression. Ans. 10*6 tons net ; 11*9 tons gross. 

10. Find the bending stress in a locomotive coupling-rod 8 ft. long, 
2 ins. broad and 4| ins. deep. It runs at 200 revolutions per minute, the 
crank radius being 11 ins. • Ans. 2*6 tons per sq. in. 



CHAPTER VIII 

1. A tie bar 9 ins. wide and 1|^ ins. thick is curved in the plane of its 
width. If there is a total tensile load on the bar of 30 tons, and if the 
mean line of pull passes 3 ins. to one side of the geometrical axis at the 
middle of the bar, find the maximum and minimum stresses at the centre 
section of the bar. (A.M.I.C.E.) 

Ans. 6f tons per sq. in. tension ; 2f tons per sq. in. compression. 

2. An upright timber post 12 ins. in diameter supports a vertical load 
of 18 tons, 3 ins. from the vertical axis of the post. Determine the maxi- 
mum and minimum stresses on a normal cross section and show by a 
diagram how the intensity of stress varies across the section. 

Ans. '477 and, '159 ton per sq. in. 

3. A cast-iron post 12 ins. in external diameter and 10 ins. internal 
diameter carries an axial load of 40 tons and also an eccentric load of 5 tons, 
parallel to the axis at an eccentricity of 12 ins. Find the maximum stress. 

■Ans. 1*98 tons per sq. in. 

4. A short wooden pillar is 20 ins. high, and rectangular in cross section, 
the thickness of the section is 6 ins., and the width 12 ins. Two vertical 
loads act on the top of the pillar, both loads act in the middle of the thick- 
ness, one of them, W^, acts at a point 1|- ins. on one side of the centre, and 
the other, Wg, acts at a point 2| ins. on the other side of the centre. If the 
stress over the base of the pillar is everywhere compressive and varies 
uniformly, its intensity being twice as great at the 6 in. edge near the line 
of action of W2 as it is at the 6 in. edge near the line of action of Wj, what 
is the ratio of W^ to Wj ? (B.Sc. Lond.) Ans. 13:11. 

5. A reinforced concrete beam, 8 ins. X 11 ins. deep, has four ^ in. 
bars, with centres at 1 in. from the bottom. Calculate for a span of 12 ft. 
the safe load (a) on the modified beam formulae; (6) on the no-tension, 
straight-line formulae. Take t = 15,000, c = 100, t, = 500, m = 15. 

Ans. (a) 1,205 lbs. ; (6) 3,920 lbs., including weight of beam. 
PP 



578 EXERCISES 

6. A reinforced concrete T beam has a flange 4 ft. X 3| in., the width of 
web being 10 ins. If the centre of reinforcement is 15 ins. below the top, 
calculate its necessary area, using the above figures. Ans. 7" 94 sq. ins. 

7. Find the relation between the depth of slab and effective depth of 
a T beam in terms of the stresses and reinforcement for the neutral axis to 
curve at the bottom of the slab. . d,\ 2 r t 

d / c 

8. A T beam is required to carry a B.M. of 320,000 in. lbs. The deptli 
to centre of reinforcement is 16 ins., and the depth of slab is 4 ins. If 
c = 600 and t — 16,000, what area of reinforcement and effective breadth 
of slab would you use ? Ans. 1'39 sq. ins. ; 12| ins. 

9. A reinforced concrete floor is 9 ins. thick, the centre of the reinforce- 
ment being 2 ins. from the bottom edge. If c = 600, t = 15,000, and 
m = 15, calculate the reinforcement necessary, and the load can that be 
safely carried. Ans. "63 sq. in. per ft. width ; 386 lbs. per sq. ft. 

10. A beam of rectangular section of breadth one half the depth is bent by 
a couple in a plane at 45° to the axes of the section. Find the safe B.M. 
in terms of those about the principal axes. ^ 9a. /^ 7 \ / ^ 



CHAPTER IX 

1. If two precisely similar beams of rectangular section, one of cast iron 
and the other of wrought iron, w^ere laid across the same span and loaded 
with the same load (within the elastic limit), what would be the relative 
deflections of the two beams? (A.M.I.C.E.) 

Ans. As E,. : E„ = about 8 : 13. 

2. A beam is of 20 ft. span and tlie movement of inertia of its section 
is 300 in. units ; what will be the central deflection for a uniformly dis- 
tributed load of 16 tons ? (A.M.I.C.E.) Ans. -72 i7i. 

3. A beam of cast iron, 1 in. broad and 2 ins. deep, is tested upon supports ' 
3 ft. apart, and shows a deflection of ^ in. under a central load of 1 ton. 
Calculate the modulus E. (A.M.I.C.E.) Aiis. 5,832 tons per sq. in. 

4. Suppose that three beams or planks, A, B, and C, of the same material 
are laid side by side across a span L = 100 ins., and a load W = 600 lbs. 
is laid across them at the centre of the span so that they all bend together. 
The beams are all 6 ins. wide, but two are 3 ins. and one 6 ins. deep. What 
will be the load carried by each beam, and what will be the extreme fibre 
stress in each ? (A.M.I.C.E.) 

Ans. 480 lbs., 60 lbs. ; 1,333 lbs. per sq. in., 667 lbs. per sq. in. 

5. Calculate the least radius to which a 1 in. round bar of wrought iron 
[E = 28 X 10^ lbs. per sq. in.l may be bent, in order that the skin stress 



EXERCISES • 579 

may not exceed 15 tons per sq. in. What is then the moment of resistance 
of the section? (A.M.I.C.E.) Ans. 34-7 ft. ; 1-47 in. tons. 

6. A beam of uniformly rectangular section is supported freely at the 
ends and carries a uniformly distributed load. Find the ratio of depth to 
span so that when the maximum stress at the centre section due to bending 
is 4 tons per sq. in,, the deflection at the centre is ^^^ of the span. 
E = 12,000 tons per sq. in. (B.Sc. Lond.) Ans. Span = 24 X depth. 

7. Find the greatest deflection in inches of a rectangular wooden beam 
carrying a load of 2 tons at the centre of a span of 20 ft., with a limiting 
intensity of stress of 1000 lbs. per sq. in. The depth of the beam is 14 ins. 
Calculate the breadth. E = 6000 tons per sq. in. (A.M.I.C.E.) 

Ans. ^ in. nearly ; 8' 2 ins. wide. 

8. A 16 in. X 6 in. x 62 lb. R.S.J, carries a load of 12 tons at quarter 
span, the span being 24 ft. Find graphically the maximum deflection and 
compare that calculated for the same beam with the load at the centre. 
(I for this section = 725*7 in. units, E = 12,500 tons/in.^) 

Ans. '46 in. ; "66 in. at centre. 

9. A simply-supported beam of uniform section and 30 ft. span is found 
to deflect 6 ins. under its own weight. Find the slope of the beam at the 
supports and also the slope which would arise if the same deflection were 
caused by a central load instead of a uniform one. Ans. '0533 ; '05. 

10. A vertical post, 24 ft. in height, supports at its upper end a horizontal 
arm projecting 6 ft. from the post. Find the horizontal and vertical 
displacements of the free end of the horizontal arm when a load of 6000 lbs. 
is suspended from it. E for post and arm = 28 X 10^ lbs. per sq. in. ; 
I for post = 412, for arm = 360 (inch units). Neglect direct compression 
of the post. (B.Sc. Lond.) Ans. Horizontal 1*55; vertical '85 in. 

11. A cantilever of circular section is of constant diameter from the 
fixed end to the middle, and of half that diameter from the middle to the 
free end. Estimate the deflection at the free end due to a weight W there. 

OQ WJ 73 

Ans. " — , where I is that at fixed end. 

24 E I 

12. A timber beam 30 ft. long and 12 ins. square in cross section rests 
on a support at each end. If a load of 1 ton is placed in the centre of the 
beam, find the work done in deflecting it. Ans. 705*6 in. lbs. 



CHAPTER X 

1. A cast-iron column has its ends securely built in. It is 12 ins. in 
external diameter, and 18 ft. long. What total load could you place on 
it if the factor of safety is 10, and the thickness of metal If ins. ? The 
constant for the Gordon formula is ^u- (B.Sc. Lond.) Ans. 163 tons. 

2. A mild steel strut, rectangular in cross section, the breadth being 



680 EXERCISES 

four times its thickness, is 9 ft. long, and has pin ends. Determine the 
cross section for 24 tons, and a factor of safety of 5. Use Rankine's 
formula, and take /,, = 67,000 lbs. per sq. in., and the constant ^^j^-^. 
(B.Sc. Lond.) 

3. Which would carry the heavier load for jBxed ends : {a) a solid mild- 
steel column 9 ins. diam. ; (6) a built-up mild-steel stanchion consisting 
of two 14x61 beams, at 8| ins. centres, with two 16 x |- in. plates each 
side ? Length in each case 14 ft. A7is. The built-up one. 

4. Discuss the formula of Gordon and Rankine in connection with the 
buckling of struts of moderate lengths, and state its limiting conditions. 
Four wrought-iron struts, rigidly held at the ends, all of section 1 in. X 1 in., 
and of lengths 15'0, 30'0, 60*0, and 90'0 ins. respectively, are found to 
buckle under loads of 15'9, 11*3, 7*7, and 4'35 tons. Test whether these 
satisfy the formula quoted, and, if so, find average values of the two 
empirical constants involved. (B.Sc. Lond.) 

5. A stanchion for a workshop has to carry a small stanchion 10 ft. long 
from the roof which carries 5 tons, and also the girder for a 15-ton crane. 
If the centre line of the roof load and crane girder are 13 ins. apart, design 
a suitable section for the stanchion. 

Ans. Two 10 ins. X 5 ins. X 30 I beams 13 ins. apart. 

6. A hollow cylindrical steel strut has to be designed for the following 
conditions. Length 6 ft., axial load 12 tons, ratio of internal to external 
diameter == '8, factor of safety, 10. Determine the necessary external 
diameter of the strut and thickness of the metal if the ends are securely 
fixed in. Use Rankine's formula, taking/ = 21 tons per sq. in., constant 
for rounded ends = y-V^ Ans. 4^ ins. diam. ; | in. thick. 

7. A steel column is built up of two 10 X 3| ins. X 28'21 lbs. channel 
sections placed 4| ins. apart, and two 12 X I in. plates at each end. If the 
ends are pin-jointed, what would you consider a safe load on a length of 
22 ft. ? Ans. 122 tons. 

8. What would be the safe load on the column of Question 7 if the load 
were 3 ins. out of centre ? Ans. 48*8 tons. 

9. Find what thickness a hollow circular cast-iron column should have 
for an axial load of 60 tons, the factor of safety being 8. The column is 
20 ft. long and is securely fixed at each end. Ans. 2 ins. 



CHAPTER XI 

1. A shaft 3 ins. in diameter, running at 250 revolutions per minute, 
transmits 50 H.P. Find the maximum stress and the twist of the shaft 
in degrees in a length of 100 ft. Take G = 12 X 10« lbs. per sq. in. 

Ans. 2,380 lbs. per sq. in. ; 28*5^ 



EXERCISES 581 

2. A turbine is connected to a dynamo placed vertically above it by a 
shaft, 2 ft. in diameter, made of steel plate | in. thick. Calculate the 
diameter of solid shaft required to transmit the same power at the same 
speed with the same maximum stress due to twist. Find the relative 
weights. (A.M.I.C.E.) Ans. ISSQ i7is. diameter ; -304:1. 

3. If a rod J in. in diameter and 20 ins. long is fixed at one end and 
the other end is twisted through an angle of 15° relatively to it, what is 
the unital strain in the outer fibres of the rod ? Ans. '00164. 

4. A bar of iron, ^ in. diameter, is twisted to destruction. Calculate 
what twisting movement is required for this purpose assuming that the 
shearing stress becomes uniform over the whole section and equals in the 
limit 19 tons per sq. in. (A.M.I.C.E.) Ans. '622 in. tons. 

5. Through what angle will a 2| inch steel shaft be twisted if it is 80 ft. 
long and the twisting movement is 19,000 in. lbs. ? Ans. 23° nearly, 

6. If the end of a rod 1 in. in diameter is twisted by the turning effort 
of a force of 80 lbs. acting at the end of a 12 in. lever, find the force which, 
when applied to the end of the same lever, would twist equally a rod of 
the same material, but of 1| ins. diameter and half the length. 

Ans. 810 lbs. 

7. A bar of mild steel 1 in. in diameter twists through an angle of 
2*2 degrees in a 20 in. length when subjected to a torque of 2,200 in. lbs. 
An exactly similar bar of the same material deflects "03 in. when loaded 
at the centre of a 20 in. simply-supported span with a load of 264 lbs. 
Calculate the value of Young's Modulus, Rigidity Modulus, Bulk Modulus 
and Poisson's Ratio. (B.Sc. Lond.) 

Ans. E = 29*88, G = 11*67, K = 32*62 million 'pounds per sq. in. 
n = *28. 

8. A shaft which runs at 135 revolutions per minute transmits 50 H.P. 
and is subjected to a bending moment equal to *75 of the twisting moment. 
What diameter of shaft is necessary on the principal stress theory ? 

Ans. 2|- ins. 



CHAPTER XII 

1. A steel wire |- in. in diameter is coiled into a spiral spring 5 ins. 
mean diameter. What weight could such a spring carry to produce a 
maximum stress in the wire of 5 tons per sq. in. ? (A.M.I.C.E.) 

Ans. 5*45 tons. 

2. Find the pull required to cause a deflection of 1 in. in a closely- 
wound helical spring of 2*5 in. mean diameter made of 120 turns of 
J inch round wire, taking G = 12 X 10^ lbs. per sq. in. Ans. 3-| lbs. 

3. A helical' spring of 3 ins. diameter is composed of 20 turns of steel 
wire '258 in. diameter. If a load of 25 lbs. is hung on it what will the 
deflection and maximum stress ? Ans. 3*18 ins. ; 5,560 lbs. per sq. in. 



582 EXERCISES 

4. A laminated spring composed of 20 plates each f in. thick and 2*95 
ins. wide has a span of 3 ft. Find the deflection under a load of 5 tons 
if E is 12,000 tons per sq. in. Ans. 2' 37 ins. 

5. How many plates f inch thick and 3 ins. broad, the largest being 
30 ins. long, are required in a leaf spring whose maximum stress is to 
be 30,000 lbs. per sq. in. with a load of 1 ton ? Ans. 8. 

6. A steel clock spring J in. wide and -^ in. thick is wound on a spindle 
fff in. in diameter. If the safe stress is 48,000 lbs. per sq. in. what is the 
maximum moment available for driving the clock ? 

Ans. 2 i/i. lbs. approx. 



CHAPTER XV 

1. A girder 100 ft. long is supported at each end and in the middle, 
and carries a uniform load of 2 tons per ft. run. Draw the B.M. and 
shear diagrams, and find the pressure on each support. (A.M.I.C.E.) 

Alls. Max. B.M. 625 ft. tons.; reaction 37*5; 125; 37'5 tons. 

2. A continuous girder consists of four spans, the two outer spans 
are each 20 ft. long, and the two inner spans are each 4(> ft. long; the 
girder carries a uniformly distributed load of H tons per ft. run. Find 
(a) The reactions at each of the piers; (b) The bending moment and shear 
at each of the piers ; (c) The position of the points of zero bending moment. 
Sketch complete bending moment and shear diagrams for the girder. 
(B.Sc. Lond.) 

3. A balk of timber, 30 ft. long, rests on two end supports, and is 
supported also by a prop which acts at a point 12 ft. from the left-hand 
end. If the balk of timber carries (including its own weight) a load of 
2 cwt. per ft. run, and if the tops of the three supports are level, determine 
the reactions at the three supports, and the bending moment at the point 
at which the prop is applied. Draw complete l^ending moment and shear 
diagram. (B.Sc. Lond.) 

4. A horizontal girder of uniform section 25 ft. lonL' is firmly fixed 
at one end, and supported by a column at 18 ft, from the fixed end. The 
girder carries a uniform load of 2 tons per ft. run of its length, and, in 
addition, a concentrated load of 30 tons at 14 ft. from the fixed end. 
When unloaded, the girder just touches, but does not exert any pressure 
on the supporting colimin. Find the pressure on the column, and draw 
bending moment and shearing force diagrams for the girder. (B.Sc. 
Lond.) 

5. A beam of 20 ft, span is built in at one end a, and is freely supported 
at other end b. It carries a imif orm load of h ton per ft. run, and a central 
isolated load of 10 tons. Draw the bending moment diagram, first finding 
the bending at end a, and show where the maximum intermediate bending 
moment occurs. Draw also the shear diagram. 



EXERCISES 583 

6. A continued girder of 2 spans, 20 ft. and 10 ft., has an overhang 
of 5 ft. from the smaller span. It carries a uniformly distributed load 
of ^ ton per ft. run, and an isolated load of 1| tons at the free end (d). 
Find the support moments, and draw the shear and B.M. diagrams. 
Determine whether this arrangement is stronger than that in which the 
support c comes below the point d. 

Arts. Max. B.M. 16*77 ft. tons ; not so strong. 

7. A beam of span I is fixed horizontally at both ends. Two equal 
loads W are placed at equal distance h from the ends of the beam. Prove 

that the greatest deflection of the beam is equal to — -— - (3 Z — 4 A), and 
^ ^ 24 E I ^ ^ 

that the bending moment at the centre of the beam is equal to — - — 

V 

(B.Sc. Lond.) 

8. A beam of 20 ft. span is fixed at the end and carries a uniformly- 
distributed load of 1 ton per ft. run from one abutment to the centre. 
Find the end B.M.s. Ans. 10-4, 22*9 ft. tons. 

9. In a continuous beam of three spans, the centre span is 72 ft. and 
the end spans 36 ft. each. A dead load of ^ ton per ft. run covers the 
whole span. Determine the support moments when a live load of 1 ton 
per ft. run covers (a) the first span ; (6) the first two spans; (c) the whole 
beam. Ans. (a) 243, 162; (6) 567, 486; (c) 547 ft. tons. 



CHAPTER XVI 

1. A C.I. beam has the following section: top flange, 4 x 1-| ins.; 
web, 12 X If ins. ; bottom flange, 12x2 ins. The centroid of the 
section is 5*5 ins. from the base of the bottom flange, and the moment 
of inertia of the section about a line through its centroid, at right angles 
to the depth, is 1200 ins.*. Draw a curve showing the intensity of shear 
at all points of the section, and find the ratio of maximum to mean shear 
stress. What proportion of shearing force is carried by the web ? (B.Sc. 
Lond. ) 

2. Find the greatest intensity of shear stress at a section of an I beam 
at which the total shear is 15 tons; the overall depth is 8 ins. ; flanges, 
6 ins. X -61 in.; web, -44 in. thick; I = 111-6 in.^. (B.Sc. Lond.) 

3. Find the ratio of the maximum to the mean shear stress on the 
section of a cast-iron beam of the following dimensions : Top flange, 
2 X ^ ins.; bottom flange, 6x1^ ins.; web, 7x1 ins. Aiis. 2'46. 

4. A beam of uniform rectangular section, 6 ins. broad by 12 ins. deep, 
is supported at the ends, and has a span of 12 feet. It carries a uniformly 
distributed load of 20 tons. At a point in the cross section, 4 feet from 



584 EXERCISES 

one end and 3 ins. vertically above the neutral axis, calculate the maximum 
intensity of compressive stress. Ans. I'll tons per sq. in. 

5. An I beam, 20 ins. deep, has flanges l^ ins. wide and 1 in. 
thick, and a web If in. thick. Tlie greatest moment of inertia is 1,650 
in. units and the total shear over the section is 80 tons. Show by a 
diagram the intensity of shear stress at all points of the section. 
(A.M.I.C.E.) Ans. Stress at N.A. — 8'44 tons per sq. in. 

6. Allowing a bending stress of 1,500 lbs. per sq. in. and a shearing 
stress with the grain of 120 lbs. per sq. in., what uniform load can be 
carried by a timber beam 12 ins. deep, 3 ins. wide and of 12 ft. span ? 

Ans. 5,760 lbs. 

7. A plate girder 4 ft. deep over the 3|ins. X 3|ins. X |in. angles has 
at each flange one plate 16 ins. x ^ in., the web being | in. thick. Find 
the distribution of shear stress on a section at which the shearing force is 
44 tons. 

Ans. Stress at N.A. = I'Ql ; at junction of angle and web = 1*25; 
just above = "481; at bottom of flange of angle = "348; just above = '014; 
junction of angle and plate = '049 ton per sq. in. 



CHAPTER XVII 

1. Show on the Bach Theory that the maximum central concentrated 
load that can be carried by a circular plate of given thickness is independent 
of the radius of the plate. 

2. What must be the thickness of a mild steel plate covering an opening 
4 ft. square if the load is 200 lbs. per sq. ft. and the safe stress is 16,000 lbs. 
per sq. in. ? Ans. '70 on the Bach Theory. 

3. Prove that on the Bach Theory the uniform load that a square 
plate of given thickness can carry is independent of the size of the plate. 

4. A rectangular reinforced concrete slab 10 ft. by 15 ft. has to carry a 
load of 300 lbs. per sq. ft. including its own weight. For what bending 
moment would you design the reinforcement in each direction ? 

Ans. 560,000 and 162,000 in. lbs. [Ranhine). 

5. A cylinder end is 12 ins. in diameter and | in. thick. Compare the 
maximum stresses on the Bach and Grashof theories if the steam pressure 
is 100 lbs. per sq. in., the end being taken as freely supported. 

Ans. 6,400 lbs. per sq. in. {Bach), 7,800 lbs. per sq. in. {Grashof). 



CHAPTER XVIII 

1. A solid steel gun has an inside diameter of 7'5 ins. and a thickness 
of r75 ins. What is the greatest tensile stress carried by an explosion 
pressure of 10,000 lbs. per sq. in. ? Ans. 27,300 lbs. per sq. in. 



EXERCISES 585 

2. What should be the external diameter of a gun whose internal diameter 
is 3|- ins., if the explosion causes a pressure of 15,000 lbs. per sq. in. and the 
allowable stress is 30,000 lbs. per sq. in. ? Ans. 5'63 ins. 

3. A solid gun has an external diameter of 12 ins. and an internal diameter 
of 6 ins. What inside pressure ^^dll cause a hoop tension of 30,000 lbs. 
per sq. in. ? Ans. 18,000 lbs. per sq. in. 

4. Find the safe internal pressure for an hydraulic press cylinder of 
external diameter 7 ins. and internal diameter 5 ins., the maximum safe 
stress being 3,000 lbs. per sq. in. Ans. 973 Ihs. per sq. in. 

5. An hydraulic press has an external diameter of 16 ins. and an internal 
diameter of 8 ins. If the pressure is 3 tons per sq. in., find the principal 
stresses at the external and internal circumference. 

Ans. External 0, 2 {tens.); internal 3 {comp.) 5 {tens.) tons per sq. in. 

6. Find the equivalent tensile hoop stresses in the problem of the above 

question if Poisson's ratio is — Ans. 5*86; 2 tons per sq. in. 

7. Find the necessary thickness of a pipe of 8 ins. internal diameter 
subjected to an internal pressure of 520 lbs. per sq. in. Adopt the maximum 
strain theory taking tj = ^ and maximum tensile stress 10,000 lbs. per 
sq. in. Ans. '22 in. 

8. A tube whose internal and external radii are 2 and 3 ins. is hooped 
so as to cause an initial hoop compression on the inside of 18,000 lbs. per 
sq. in. What will be the tensile stress at the inside if the explosion causes 
a pressure of 25,000 lbs. per sq. in. ? Ans. 29,000 lbs. per sq. in. 



CHAPTERS XIX AND XX 

1. Show that for a curved beam of rectangular section of depth d the 
deviation of the neutral axis from the centre is given approximately by 

, R being the radius of curvature of the centre line. 
12 sx 

2. A rectangular bar 1 in. wide and 2 ins. deep is bent to a radius of 
4 ins. and is used as a hook. Find the maximum stress caused by a load 
of 1,000 lbs. acting through the centre of curvature of the bar. 

Ans. 7,700 lbs. per sq. in. 

3. Find the maximum bending moments upon a chain link made of 
f in. circular stock, the link being 5 ins. deep and 3 ins. broad. Treat the 
ends as circular and the sides as straight. The load carried is 3,000 lbs. 

Ans. 1,.300 in. lbs. ; 387 in. lbs. 

4. Find the stress due to centrifugal force in the rim of a cast-iron 
flywheel 8 ft. in diameter running at 160 revolutions per minute. 

Ans. 437 lbs. per sq. in. 



586 EXERCISES 

5. A steel shaft is of 13*4 in. diameter and is 98 ft. long. Find its 
critical speed when unloaded. Ans. 46 revolutions per minute. 

6. Find the angular velocity at which whirling will start in an unloaded 
steel shaft 3 ins. in diameter and 11 ft. long. Ans. 75 radians per second. 

7. Find the whirling speed of a shaft carrying a central load of 1,170 lbs. 
between swivelled bearings 7 ft. apart. The shaft is 3-J- ins. in diameter. 

Ans. 725 revolutions per minute. 



APPENDIX. 



The following Tables give the properties of the British Standard 
Sections which are usually listed by makers. 




-t 




Pi 



k 



\ Y 



r 



v 



t 



\ 



f\ 



Y^ V 






I 

T 



\ 



W:^ 



J 



X — 



5-1 

\ I 
r4-- -X 



::i c 



J 

-1 ^ . 



Fig. A. — Properties of British Standard Sections, 



5s: 



AppeJidix. 



BRITISH STANDARD SECTIONS.* (See Fig. A.) 
Properties of British Standard I Beams. 



Size 


\Vt. 
per 1 

foot r 


Thickness 

1 


1 
1 

II i 


Moments of 
Inertia 


i 
Section 
Moduli 


Radii of 
(jyration 


H. B. 


1 






I 


1 
1 




I 








/ 
ins. 


T 


X 


XX 


vv j 


XX 


YY 

1 


XX i 

1 
"1 

ins. 


YY 


inches. 


lb. 


ins. 1 


I 

sq. ins. ^ 


ins. 


ins. 


ins. 


ins. 


ins. 


3.x i| ■ 


4 


•16 


•248 


i-i8 


1-66 


•124 


I'll 


•165 


I-I9 


•325 


3x3, 


8| 


•20 


•332 


2-50 


379 


1-26 


2-53 


•841 


1-23 


710 


4 X i| 


5 


•17 


•240 


1-47 


3-67 


•194 


1-84 


•222 


1-58 


•363 


4x3; 

1 


9* 


•22 


•336 


2 -80 


7-53 


1-28 


376 


•854 


1*64 


•677 


4|x if| 


6| 


•18 


..23 


1-91 ' 


677 


•263 


2-85 


•300 


1-88 


•371 


5 X 3 


1 1 


•22 


•376 


3-24 1 


13-6 


I "46 


5*45 


•974 


2-05 


•672 


5 X 4i 


18 


•29 


•448 


5-29 


227 


5-66 


9-08 


2-51 


2-07 


1-03 


6x3 


12 


•26 


•348 


3-53 


20*2 


I '34 


674 


•892 


2*40 


•616 


6 X 4| 


20 


'Zl 


•431 


5*88 


347 


5*41 


11-6 


2*40 


2-43 


•959 


6x5 


25 


•41 


•520 


7-35 


43*6 


9-II 


14-5 


3*64 


2*44 


rii 


7x4 


16 


'^-':> 


•387 


471 


39-2 


3*41 


I I -2 


171 


2-89 


•851 


8x4 


18 


•28 


•402 


5-30 


557 


3'57 


13-9 


179 


3*24 


•821 


8x5 


28 


•35 


•575 


8-24 


89-4 


10-3 


22'3 


4-IO 


3-29 


1*12 


8x6 


35 


•44 


•597 


10-3 


II I 


17-9 


27-6 


5-98 


3-28 


1-32 


9x4 


21 


•30 


•460 


6-i8 


8i-i 


4'20 


i8-o 


2*IO 


3-62 


•824 


9x7 


58 


•55 


•924 


17T 


230 


46-3 


51-1 


13-2 


3-67 


1-65 


10 X 5 


30 


•36 


•552 


8-82 


146 


978 


29'I 


3-91 


4 "06 


1-05 


10 X 6 


42 


•40 


•736 


12-4 


212 


22*9 


42-3 


7-64 


4-14 


1-36 


10 X 8 


70 


•60 


•970 


20'6 


345 


71-6 


69*0 


17-9 


4-09 


1-87 


12x5 


32 


•35 


•550 


9-41 


220 


974 


367 


3-90 


4-84 


I -02 


12x6 


44 


•40 


•717 


12*9 


315 


22 '3 


52-6 


7 "42 


4*94 


1-31 


12x6 


54 


•50 


•883 


15-9 


376 


28-3 


62-6 


9 -43* 


4-86 


1-33 


14x6 


46 


•40 


•698 


13*5 


441 


21 "6 


62-9 


7 -20 


571 


1-26 


14x6 


57 


•50 


•873 


i6-8 


553 " 


27-9 


76-2 


9"3i 


5-64 


1-29 


15 X 5 


42 


•42 


•647 


12-4 


428 


II -o 


57-1 


478 


5-89 


•983 


15x6 


59 


•50 


•880 


17-3 


629 


28-2 


83-9 


9-40 


6 '02 


1-28 


16 X 6 


62 


*55 


•847 


1 8-2 


726 


27-1 


907 


9*02 


6-31 


1-22 


18 X 7 


75 


1 

'55 


•928 


22*1 


1 1 50 


46-6 


128 


^ys 


7-22 


I '45 


20 X 7I 


89 


' -Go 


i-oi 


26*2 


1671 


62-6 


167 


167 


7 '99 


1-55 


24 X r\ 


100 


1 -60 
1 


lro7 


29-4 


2655 


66-9 


221 


17-8 


9-50 


1-51 



• Published by permission of the Engineeniig Standards Committee. The Tables of British 
Standard I Beams, Channels, and Zed Bars are reprinted from Report No. 6 as issued by the Committee. 
Additional calculations have been inserted in the Tables of British Standard Unequal Angles, Equal 
Angles, and Tee Bars for thicknesses other than those calculated by the Coir'»iittee, such calculations 
having been taken by permission from the Pocket Companion issued by Messrs. Dorman, Long & Co., Ltd. 

588 



Appendix, 
Properties of British Standard Channels. 



Size 


Standard 
Thicknesses 


Weight 
per 
foot 


Area 


c 


H 

5 


Moments of 
Inertia 


Section 
Moduli 


Radii of 
Gyration 


A X B 


t T 

1 


About 
XX 


About 
YY 


About 
XX 


About 
YY 


About 
XX 

ins. 


About 
YY 


ins. 


ins. 


ins. 


lbs. 


sq. ins. 


ins. 


ins. 


ins. 


ins. 


ins. 


ins. 


15x4 


•525 


•630 


41-94 


12-334 


•935 


377-0 


14-55 


50-27 


4-748 


5-53 


1-09 


12x4 


•525 


•625 


36-47 


10727 


I -03 1 


218-2 


13-65 


36-36 


4*599 


4-51 


I-13 


I2X3i 


•500 


•600 


32-88 


9-671 


-867 


190-7 


8-922 


31-79 


3-389 


4*44 


-960 


12x3^ 


•375 


•500 


26'IO 


7-675 


-860 


158-6 


7-572 


26-44 


2-868 


4*55 


■993 


11x3-2 


•475 


•575 


29-82 


8-771 


•896 


148-6 


8-421 


27-02 


3-234 


4-12 


-980 


10x4 


•475 


•575 


30-16 


8-871 


I-I02 


1307 


12-02 


26-14 


4*147 


3*84 


I-16 


10 X il 


•475 


•575 


28-21 


8-296 


■933 


117-9 


8-194 


23-59 


3-192 


3'77 


*994 


10x3^ 


•375 


•500 


23-55 


6-925 


•933 


102-6 


7-187 


20-52 


2 -800 


3*85 


I -02 


9x3! 


•450 


•550 


25*39 


7-469 


-971 


88-07 


7-660 


19-57 


3-029 


3*43 


I -01 


9x35 


•375 


•500 


22-27 


6-550 


•976 


79-90 


6-963 


17-76 


2-759 


3*49 


1-03 


9x3 


•375 


•437 


19-37 


5-696 


•754 


65-18 


4-021 


14-48 


1-790 


3*38 


-840 


8x3* 


•425 


•525 


22-72 


6-682 


I -Oil 


63-76 


7-067 


15-94 


2-839 


3-09 


1-03 


8x3 


•375 


•500 


I9-3C 


5 '67 5 


•844 


53*43 


4-329 


13-36 


2-008 


3*07 


'S73 


7x3! 


•400 


•500 


20-23 


5 "950 


I -061 


44-55 


6-498 


12-73 


2-664 


274 


1*04 


IX-:, 


•375 


•475 


17-56 


5-166 


-874 


37-63 


4-017 


10-75 


1-889 


2-70 


-882 


6x3i 


•375 


•475 


17-9 


5-266 


1-119 


29-66 


5-907 


9-885 


2-481 


2-36 


1-06 


6x3 


•312 


•437 


14-49 

1 


4-261 


-938 


24-01 


3-503 


8-003 


1-699 


2-37 


-907 



Properties of British Standard Zed Bars. 



Size 


Standard 
Thicknesses 


Area 


Weight 
per 
foot 


Moments of Inertia 


Section 


Moduli 


c 

■~ in 

c -a 

< 


1 "2 
-5.2 

4-1 >^ 


A X B 


/ 


T 


About 
XX 


About 
YY 


About 
XX 


About 
YY 




ins. 


ins. 


ins. 


sq. ins. 


lbs. 


ins. 


ins. 


ins. 


ins. 




ins. 


10 X 3^ 


*475 


•575 


8-283 


28-16 


117-865 


12-876 


23-573 


3-947 


14 


-839 


9x3! 


-450 


-550 


7-449 


25-33 


87-889 


12-418 


19-531 


3-792 


i6i 


*843 


8x31 


-425 


-525 


6-670 


22-68 


63729 


12*024 


15-932 


3-657 


i9i 


-845 


7x31 


•400 


-500 


5-948 


20-22 


44-609 


ii-6i8 


12-745 


3-521 


23 


-840 


6X3^ 


•375 


;475 


5-258 


17-88 


29-660 


11-134 


9-887 


3-361 


28i 


•821 


5x3 


•350 


^-450 


4-169 


14-17 


16-145 


6-578 


6-458 


2-328 


29i 


-698 



589 



Appendix. 
Properties of British Standard Unequal Angles. 



Size and 






Weight 


Dimensions 


Moments of 
Inertia 


Section 
Moduli 






- 
Thickness 


Area 


per 














01 u. 


>> 






















foot 


J 


P 


About 


About 


About 


About 


<y 










XX 


Y Y 


XX 

ins. 


YY 




►J ° 


ins. 




sq. ins. 


lb. 


ins. 


ins. 


ins. 


ins. 


ins. 


ins. 


1 y^ }k^ 


i 


5'o 


17-00 


2-50 


-764 


25-1 


4-28 


5-58 


1-56 


I4| 


-74 


5) ?» 


5 

s 


6'i72 


20-98 


2-55 


•814 


30-55 


5-15 


6-86 


1-92 


I4i 


-74 


M ?1 


1 


7-313 


24-86 


2-60 


•862 


35-68 


5-95 


8-II 


2-26 


14 


•73 


6i X 4| X 


1 
2 


5-248 


17-84 


2-08 


1-09 


22 "1 


8-75 


5-02 


2-57 


25 


'97 


11 11 


5 

8 


6-482 


22-04 


2-13 


I-14 


27*09 


10-60 


6-20 


3-15 


25 


-96 


11 11 


3 

4 


7-686 


26-13 


2-i8 


I-19 


31-66 


12 -32 


Ty:> 


3-72 


25 


•96 


6| X 3i X 


3 

8 


3-610 


12-27 


2-22 


-741 


15-7 


3-27 


y(^i 


I-18 


16^ 


'75 


ii 11 


J. 
2 


4-750 


16-15 


2-28 


•792 


20-4 


4-20 


4-83 


1-55 


16^ 


•75 


j» 'J 


5 

s 


5-860 


19-92 


2*33 


-841 


24-83 


5-06 


5-95 


1-90 


16 


•74 


6 X 4 X 


3 

s 


3-610 


12-27 


I-9I 


•923 


13-2 


4-73 


3-23 


1-54 


23i 


-87 


11 j» 


1 
2 


4-750 


16-15 


1-96 


•974 


17-1 


6-10 


4-23 


2-02 


23i 


•86 


11 ii 


5 


5-860 


19-92 


2 -02 


I -02 


20-8 


7-36 


5-23 


2-47 


23I 


•86 


6 X 3| X 


3, 


3-424 


11-64 


2-01 


ll-':> 


12-6 


3-22 


3-16 


1-18 


19 


•76 


" n 


1 


4-502 


15-31 


2-06 


•823 


16-4 


4-14 


4-16 


1-55 


19 


•75 


1' ?j 


5 

8 


5-549 


18-87 


2-II 


•872 


19-88 


4-97 


5-II 


1-89 


18^ 


•75 


5! >' 3i X 


3 

S 


3-236 


1 I -00 


1-80 


-807 


9-93 


3-15 


2-68 


1-17 


22 


•76 


M -5 


\ 


4-252 


14-46 


1-85 


-857 


12-80 


4-05 


3-51 


i'53 


22 


•75 


;5 11 


5 
8 


5-236 


17-80 


1-90 


-905 


15-6 


4-86 


4 jj 


1-87 


21^ 


•75 


5i X 3 X 


3. 


3-050 


10-37 


1-90 


-662 


9-45 


2-02 


2-62 


•86 


17 


-64 


J? ?> 


1 

2 


4-003 


13-61 


1-95 


-711 


12-2 


2-58 


3 "44 


1-13 


16^ 


-64 


?? V 


5 

8 


4-925 


16-74 


2-00 


-759 


14-7 


3-08 


4-20 


XV 


\b\ 


•63 


5 X 4 X 


3 

8 


3-236 


II -oo 


1-51 


i-oi 


7-96 


4-53 


2-28 


1-52 


32 


-85 


» 5? 


4 


4-252 


14-46 


1-56 


1-06 


10-3 


5-82 


2-99 


1-98 


32 


•84 


« 11 


5 

8 


5-236 


17-80 


I -60 


i-ii 


12-4 


7-OI 


3-66 


2-43 


32 


-83 


5 X 3^ X 


3 

8 


3-050 


10-37 


1-59 


-848 


7-64 


3-09 


2-24 


I-I7 


25^ 


•75 


» 5> 


1 
2 


4-003 


13-61 


1-64 


•897 


9-86 


3-96 


2-93 


1-52 


25i 


•75 


)5 M 


5 
8 


4-925 


16-74 


1-69 


-944 


II-9 


4-75 


3-60 


1-86 


25 


•74 


5 X 3 X 


Vb 


2-402 


8-17 


1-66 


•667 


6-14 


1-68 


1-84 


•72 


20 


•65 


5) 5> 


3 

8 


2-859 


! 9-72 


1-68 


-693 


7-24 


1-97 


2-i8 


-85 


i9i 


•65 


)) V 


1 
2 


3"749 


12-75 




■742 9*33 


2-51 


2-85 


i-ii 


19^ 


•64 


J: 35 


5 

8 


4-609 


15-67 

1 


1-78 


•789 11-25 


3-00 


3-49 


1-36 


19 


-64 



590 



Appendix, 
Unequal Angles {continued). 



Size and 

Thickness 



4^ X 



ins. 
1>\ X 



4 X 



J 2 



4 X 



X -\ 13 



32 X 



Z\ X 2; 



8 

\ 
2 

5. 

8 

X 16 



X 2.^ X 



X 2 



Area 



I 
I 

{Weight 

i per 

foot 



sq. in. 
2 "402 

2-S59 

3749 
4*609 

2'246 

2'67i 

3 '499 
4-295 
2*091 
2-485 
3-251 

3-985 
I '934 
2*298 
3-001 

3*673 
1-799 
2*111 
2-752 
1-312 
1*921 



2^ X 



2 X 



)» 




2 


2-499 





X 


I 


1*187 


?) 




3 

6 


^'733 


55 




1 
2 


2*249 


2 


X 


1 
4 


1*063 


55 




I'e 


1*309 


V 




s 


1*547 


li 


X 


1% 


-622 


51 




1 

4 


•814 


55 




I'fi 


•997 



lb. 

8-17 

9-72 

12*75 

15*67 

7*64 

9*o8 

11*90 

14*61 

7*11 

8*45 
11*05 

13-55 
6-58 

7-8i 
10-20 
12*49 
6-05 
7-18 
9-36 
4-46 

O DO 
8*50 
4*04 
5-89 
7-65 

3-6i 

4-45 
5*26 
2*11 

2*77 

3-39 



Dimensions 



ins. 
-36 
-39 
-44 
-48 
*i6 



ins. 

*866 
*89i 

-940 
•987 
-915 



Moments of 
Inertia 



About 
X X 



1*19 


•941 


1*24 


•990 


1-28 


1*04 


1*24 


•746 


1*27 


*77i 


1*31 


*8i9 


1*36 


*865 


1*04 


*792 


1*07 


■*8i9 


I*II 


-867 


1*16 


-912 


1*12 


*627 


1*15 


*652 


1-20 


-699 


•895 


*648 


-945 


-6,7 


•992 


-744 


-976 


-482 


1*03, 


•532 


1*07 


*578 


'774 


-527 


•799 


•552 


•823 


•575 


•627 


•381 


-653 


•407 


*678 


-431 



ins. 
4*22 
5-69 
7*31 

8*8r 
3-46 
4*08 

5-23 
6-28 

3*31 
3-89 
4-98 
5-96 
2*27 
2*67 
3-40 
4*05 
2*15 
2*52 
3-20 
1*14 
I -62 
2-05 
I -06 
1*50 
1-89 
■636 
-770 
•895 
-240 
•308 
•369 



About 
Y Y 



ms. 
2*55 
3*00 

3-84 
4*61 

2*47 
2*90 

371 
4-44 
1*59 

1-87 

2*37 
2*83 

1-53 
1*80 

2*28 
2*71 
•910 
I -06 

1-34 
*7i6 
1*02 
1-28 

'373 
*525 
*656 
•359 
•433 
•502 
*ii5 
'1 46 
-174 



Section 
Moduli 



About 
X X 



ins. 

1-54 
1-83 
2-39 
2-92 

1*22 

1-45 
1-89 
2*31 
1*20 

1-42 
1-85 
2-26 
*92 
1*10 
1*42 

i-73 

-90 

I -07 

■-39 
-54 
-79 

1*02 
•52 
•76 
-98 
'37 
*45 
-53 
•17 
•23 



About 
Y Y 



*97 
1-15 

1*5 

1*83 

-96 

1*13 

1*48 

1-80 

•71 

•84 

I -09 

1*33 
•69 

-83 
1*07 

1-30 

-49 

•57 
•74 
*39 
-57 
'73 
-25 
-36 
-46 
•24 
•30 

•35 
•10 

*i3 
•16 



c 

•— {/I 

o f 



302 

3oh 

30 
30 

37 

37 

37 

3H 
28^ 

28^ 

28^ 

28 

35i 
3Sh 
35i 
35 

26| 

26 
26 
34 

34 

33h 

23* 

23 

22.^ 

32 
3ii 
31I 
28^ 

28 
28 



'-5.2 
ft: 2 

»^ 

QJ <^ 

J ° 

ins. 
*74 
-74 
*74 
•74 

*72 
*72 

•71 
•71 
-64 

-64 
*63 
■63 

*62 
*62 

•61 

*6i 
•54 
*53 
*53 

•52 

-52 
•52 
*43 
*42 
-42 
•42 
•42 
•42 
•32 



591 



Appendix. 



British Standard Equal Angles. 



Sizes 


Area 


Weight 
per foot 


T 
J 


1 

! 


1 

Section 

Modulus 

about X X 


Least 
Radius of 
Gyration 




ins. 




sq. ins. 


lb. 


ins. 


ins. 

1 


ins. 


I 

ins. 


8 X 


8 X 


i 


775 


26-35 


2-15 


47'4 


8-IO 


1-58 


8 X 


8 X 


5. 


9-61 


32-67 


2-20 


1 58-2 


10-03 


^'11 


8 X 


8 X 


3 
4 


11-44 


38-89 


2-25 


68-5 


11-91 


1-56 


6 X 


6 X 


I'S" 


5-06 


17-21 


1-64 


17*3 


3*97 


I-18 


6 X 


6 X 


5 

S 


7-II 


24-18 


1-71 


23-8 


5-55 


I-18 


6 X 


6 X 


3 

4 


8-44 


28-70 


1-76 


27-8 


6-56 


I-I7 


5 X 


5 X 


3 

S 


3-6i 


12*27 


1-37 


8-51 


2-24 


'98 


5 X 


5 X 


1 


475 


16-15 


1-42 


I i-o 


3-07 


•98 


5 X 


5 X 


5 

S 


5-86 


19-92 


1-47 


13-4 


3-80 


-98 


4ix 


4ix 


3 

S' 


3'24 


I I -oo 


1 -22 


6-14 


1-87 


-88 


4|x 


45 X 


1 
2 


4-25 


14-46 


1-29 


7-92 


2-47 


-87 


45 X 


45 X 


5. 

S 


5-24 


17-80 


I '34 


9-56 


3-03 


'^1 


4 X 


4 X 


3 

S 


2-86 


972 


1-12 


4-26 


1-48 


78 


4 X 


4 X 


1 


375 


12-75 


I-I7 


5-46 


1-93 : 


'17 


4 X 


4 X 


5 

S 


4-6i 


15-67 


I -22 


6-56 


2-36 


•77 


35 X 


3^x 


"t'« 


2*09 


7-II 


•97 


2-39 


'95 


-68 


3ix 


3ix 


3 

& 


2-48 


8-45 


I "OO 


2-80 


1*12 


•68 


35 X 


3^x 


1 
2 


3-25 


11-05 


1-05 


3'57 


1-46 : 


-68 


35 X 


3l X 


5 

S 


3-98 


i3'55 


1-09 


4-27 


177 


-68 


3 X 


3 X 


1 
4 


1-44 


4-90 


•827 


I-2I 


•56 


•59 


3 X 


3 X 


3 

8 


2'II 


7-18 


•877 


1-72 


-81 


•58 


3 X 


3 X 


1 
2 


275 


9-36 


-924 


2-19 


1-05 


•58 


3 X 


3 X 


5 


3-36 


11-43 


•970 


2-59 


1-28 


-58 


l\ X 


2^ ><: 


1 

4- 


I-I9 


4-04 


703 


■677 


•38 , 


•48 


2^ X 


l\ X 


t''« 


1-46 


4-98 


•728 


-822 


•46 , 


•48 


2^X 


2^X 


3 

8 


173 


5-89 


752 


-962 


'55 


•48 


2| X 


2^ X 


1 


2-25 


■7-65 


799 


I-2I 


-71 


•48 


2^X 


2\ X 


16 


•809 


275 


•616 


•378 


•23 


•44 


2f X 


2} X 


1 
4 


I '06 


3-6i 


•643 


•489 


•30 


•44 


2^X 


2|X 


IB 


i'3i 


4-45 


-668 


•592 


•^:>7 


•43 


2r X 


2|X 


3 

S 


i'55 


5-26 


•692 


•686 


•44 


•43 



592 



Appendix. 
British Standard Equal Angles (continued). 



Sizes 


Area 


WViifht 
per 
foot 


J 


I.XX 


Section 

Modulus 

about X X 


Least 
Radius of 
Gyration 




ins. 




sq. ins. 


lb. 


in. 


in. 


in. 


in. 


2 X 


2 X 


i« 


715 


2-43 


•554 


•260 


-18 


•39 


2 X 


2 X 


1 

4 


-938 


3*19 


•581 


•336 


•24 


•39 


2 X 


2 X 


tV 


I-I5 


3-92 


•605 


•401 


•29 


•38 


2 X 


2 X 


3 

s 


1-36 


4-62 


•629 


•467 


•34 


•38 


ifx 


i|x 


.'5 

.1« 


•622 


2-II 


•495 


•172 


•U 


•34 


ijx 


i^x 


1 

4 


-814 


2-77 


•520 


•220 


•]8 


•34 


i|x 


if X 


5 
16 


•997 


3*39 


•544 


•264 


•22 


•34 


I^X 


I^ X 


IG 


•526 


179 


•434 


•105 


•10 


•29 


[*X 


I^X 


1 
4 


•686 


2-33 


-458 


•134 


•13 


•29 


i|x 


4x 


"16 


•839 


2-85 


-482 


•159 


•16 


-29 


I4-X 


IjX 


16 


•433 


1-47 


•371 


•058 


•07 


•24 


I^X 


\\ X 


1 
4 


•561 


1-91 


•396 


•073 


-C9 


•23 



British Standard Tees. 



SizfS 




Area 


Weight 
per foot 


J 


Moments of 
Inertia 


Section 


Moduli 


Radii of Gyration 




XX 


YY 

ins. 


XX 


YY 


XX 


YY 


ins. 




sq ins. 


lb. 


ins. 


ins. 


ins. 


ins. 


ins. 


ins. 


6 X 4 X 


3 

s 


3-634 


12-36 


-915 


4-70 


6*34 


1-52 


2-II 


I-I4 


1-32 


6 X 4 X 


1 

2 


4-771 


l6-22 


-968 


6-07 


8-62 


2-00 


2-87 


I-13 


1-34 


6 X 4 X 


5 

s 


5-878 


19-99 


I -02 


7*35 


10-91 


2-47 


3-64 


I - 1 2 


1-36 


6 X 3 X 


8 


3-260 


II -08 


•633 


2 -06 


6*39 


-87 


2-13 


-795 


1-40 


6 X 3 X 


\ 


4-272 


14-53 


•684 


2-63 


8-65 


I-14 


2-88 


-785 


1-42 


6 X 3 X 


5 

S 


5-256 


17-87 


•732 


3*14 


10-94 


1-39 


3-65 


'ir:> 


1-44 


5 X 4 X 


8 


3-257 


11-07 


-998 


4'47 


3-69 


1-49 


1-48 


I-I7 


I -06 


5 X 4 X 


1 



4-268 


14-51 


1-05 


5*77 


5-02 


1-96 


2-01 


I-I6 


1-08 


5 X 3 X 


3 

8 


2-875 


9-78 


•691 


1-97 


3*71 


-85 


1-49 


-828 


1-14 


5 X 3 X 


1 
2 


3762 


12-79 


-741 


2-52 


5-03 


i-ii 


2-01 


-818 


1-16 


4 X 4 X 


3 

8 


2-872 


9-77 


l-II 


4-19 


1-90 


1-45 


•95 


I-2I 


-814 


4 X 4 X 


1 

2 


3758 


12-78 


i-i6 


5-40 


2-59 


1-90 


1-29 


I -20 


•830 


4 X 3 X 


5. 


2-498 


8-49 


•767 


1-86 


1-91 


•83 


-96 


-863 


-875 


4 X 3 X 


h 


3-262 


11-08 


-816 


2-30 


2-60 


1-08 


1-30 


-851 


-893 



QQ 



593 



Appendix. 

British Standard Tees ( confiniied). 



Sizes 


1 

Area j 


Weight 
per 1 
foot ] 


J 


Moments of 
Inertia 


Section Moduli 


Radii of Gyratioa 


1 


XX 


YY 


XX 


YY 


XX 


YY 


:ti3. sq. ins. 


lbs. 


ins. 


I'lS. 


1 05. 


ins. 


ins. 


ins 


sns. 


3i X 3| X f 


2-496 


8*49 


-98 


2*79 


1*28 


no 


'IZ 


1*05 


•717 


3? X 3^ X ^ 


3-259 


11*08 


1*04 


3-54 


1-75 


1-44 


I'OO 


1*04 


•733 


3 X3 xf 


2*121 


7*21 


•868 


1*70 


*8i6 


*8o 


-54 


•897 


•620 


3 X3 x^ 


2*760 


9-38 


*9i8 


2*16 


i*ii 


1*04 


•74 


*886 


*636 


3 X2|xf 


1*929 


6*56 


•695 


i-oi 


•814 


-56 


•54 


•725 


■650 


3 X2^X^ 


2*506 


8*52 


•742 


1-28 


1*12 


'l?i 


•74 


•713 


•665 


2^ X 2^ X ^ 


I-I97 


4-07 


•697 


•677 


*302 


•38 


•24 


-752 


*502 


2^ X 2| X -T% 


1-474 


5*01 


•724 


•832 


•387 


•46 


•31 


•747 


*5I2 


2^ X 2|x f 


1*742 


5*92 


•750 


-959 


•473 


-55 


•38 


•742 


•521 


2ix 2^X^ 


1*071 


3-64 


•638 


*488 


*224 


-30 


•20 


■675 


•457 


2^X2^X1 


1-554 


5-28 


*689 


*685 


•349 


•44 


■31 


*664 


•474 


2 X 2 X J 


•947 


3*22 


•579 


'lyi 


•157 


•24 


•16 


-597 


-407 


2 X2 xf 


1*367 


4*64 


•628 


•469 


•246 


•34 


■25- 


•5S6 


•424 


1^X2 X J 


•820 


2*79 


*648 


•307 


•068 


-23 


*o9 


•612 


*288 


I4X2 X-,% 


1*003 


3-41 


•674 


•369 


*oS8 


*28 


•12 


■607 


•296 


ifx i|x^ 


*820 


2*79 


-519 


*22I 


•107 


•18 


•12 


•520 


•361 


i|xi|Xi% 


-999 


3-40 


•544 


•265 


•137 


•22 


*i6 


■515 


•370 


i^x iix^% 


-531 


1*81 


-435 


*io6 


*048 


•ID 


*o6 


•447 


•301 


I^X T^X^ 


*692 

1 


2-35 

1 


•460 


•135 


*o67 


-.3 


*o9 


•442 


•312 



.^94 



MATHEMATICAL TABLES 



595 



Angle 


Chord. 


Sine. 


1 

Tangent. 


Co-tangent. 


1 
Cosine. 

i 








De- 
grees 


Radians. 


0^ 














X 


1 


1-414 


1^5708 


90' 


1 
•> 

3 


■0175 
•0349 
•0524 
•0698 


•017 
•035 
•052 
•070 


•0175 
•0349 
•0523 
•0698 


•0175 
•0349 
■0524 
•0699 


57-2900 
28-6363 
19-0811 
14-3007 


-9998 

-9994 

j -9986 

-9976 


1-402 
1-389 
1-377 
1-364 


1"5533 
1-5359 
1-5184 
1-5010 


89 
88 
87 
86 


5 


•0873 


•087 


•0872 


•0875 


11-4301 


i -9962 


1-351 


1-4835 


85 


6 
7 

8 
9 


•1047 
•1222 
•1396 
•1571 


•105 
•122 
•140 
•157 


•1045 
•1219 
•1392 
•1564 


•1051 
•1228 
•1405 
•1584 


9-5144 
8-1443 
7-1154 
6-3138 


-9945 

, -9925 

•9903 

-9877 


1-338 
1-325 
1-312 
1-299 


1-4661 
1-4486 
1-4312 
1-4137 


84 
83 
82 
81 


10 


•1745 


•174 


•1736 


•1763 


5-6713 


•9848 


1-286 


1-3963 


80 


11 
12 
13 
li 


•1920 
•2094 
•2269 
•2443 


•192 
•209 
•226 
•244 


•1908 
•2079 
•2250 
•2419 


•1944 
•2126 
•2309 
•2493 


5-1446 
4-7046 
4-3315 
4-0108 


-9816 
-9781 
-9744 
-9703 


1-272 
1-259 
1-245 
1-231 


1-3788 
1-3614 
1-3439 
1-3265 


79 

78 
77 
76 


15 


•2618 


•261 


•2588 


•2679 


3-7321 


•9659 


1-218 


1-3090 


75 


16 
17 
18 
19 


•2793 
•2967 
•3142 
•3316 


•278 
•296 
•313 
•330 


•2756 
•2924 
•3090 
•3256 


•2867 
•3057 
•3249 
•3443 


3-4874 
3-2709 
3-0777 
2^9042 


-9613 
-9563 
-9511 
•9455 


1-204 
1-190 
1-176 
1-161 


1-2915 
1-2741 
1-2566 
1-2392 


74 
73 
72 
71- 


20 


•3491 


•347 


•3420 


•3640 


2^7475 


•9397 


1-147 


1-2217 


70 


21 
22 
23 
24 


•3665 
•3840 
•4014 
•4189 


•364 
•382 
•399 
•416 


•3584 
•3746 
•3907 
•4067 


•3839 
•4040 
•4245 
•4452 


2-6051 
2-4751 
2-3559 
2-2460 


1 
•9336 
-9272 
-9205 
•9135 


1-133 
1-118 
1-104 
1-089 


1-2043 
1-1868 
1-1694 
1-1519 


69 
68 
67 
66 


25 


•4363 


•433 


•4226 


•4663 


2-1445 


•9063 


1-075 


1-1345 


65 


26 
27 
28 
29 


•4538 

•4n2 

•4887 
•5061 


•450 
•467 
•484 
•501 


•4384 
•4540 
•4695 

•4848 


•4877 
•5095 
•5317 
•5543 


2-0503 
1-9626 
1-8807 
1-8040 


•8988 
•8910 
•8829 
•8746 


1-060 
1-045 
1-030 
1-015 


1-1170 
1-0996 
1-0821 
1-0647 


64 
63 
62 
61 


30 


•5236 


•518 


•5000 


•5774 


1-7321 


•8660 


1-000 


1-0472 


60 


31 
32 
33 
34 


•5411 
•5585 
•5760 
•5934 


•534 
•551 
•568 
•585 


•5150 
•5299 
•5446 
•5592 


•6009 
•6249 
•6494 
•6745 


1-6643 
1-6003 
1-5399 
1-4826 


•8572 
•8480 
•8387 
•8290 


•985 
•970 
•954 
•939 ' 


1-0297 

1-0123 

-9948 

-9774 


59 
58 
57 
56 


35 


•6109 


•601 


•5736 


•7002 


1-4281 


•8192 


•923 


-9599 


55 


36 
37 
38 
39 


■6283 
•6458 
•6632 

•6807 


•618 
•635 
•651 
•668 


•5878 
•6018 
•6157 
•6293 


•7265 
•7536 
•7813 
•8098 


1-3764 
1-3270 
1-2799 
1-2349 


•8090 
•7986 
•7880 
•7771 


•908 
•892 
•877 
•861 


-9425 
•9250 
•9076 
-8901 


54 
53 
52 
51 


40 


•6981 


•684 


•6428 


•8391 


1-1918 


•7660 


•845 


-8727 


50 


41 
42 
43 
44 


•7156 
•7330 
•7505 
•7679 


•700 
•717 
•733 
•749 


•6561 
•6691 
•6820 
•6947 


•8693 
•9004 
•9325 
•9657 


1-1504 
1-1106 
1-0724 
1-0355 


•7547 
•7431 
•7314 
•7193 


•829 
•813 
•797 
•781 


•8552 
•8378 
•8203 
•8029 


49 

48 
47 
46 


45- 


•7854 


•765 


•7071 


1-0000 


1-0000 


•7071 


•765 


•7854 


45 






Cosine 


Co-tangent 


Tangent 


Sine 


Chord 


Radians 


Degrees 


Ang 


'le 



596 



MATHEMATICAL TABLES 
Logarithms 








1 


2 


3 


4 


5 


6 


7 8 9 


12 3 4 


5 


6 7 8 9 


10 

11 

12 
13 
14 


UCKXJ 


0043 0086 

I 


0128 


0170 


0212 


0253 


0294 0334 


0374 


4 9 13 17 
4 8 12 16 


21 
20 


26 30 34 38 
24 28 32 37 


0411 


0453 0492 


0531 


0569 


0607 


0645 


1 
0682 0719 


4 8 12 15 
0755 4 7 11 15 


19 
19 


23 27 31 35 
22 26 30 33 


0792 
1139 


0828 0864 


0899 


0934 


0969 


1 
1004 1038 


1072 


1106 


Is 7 11 14 
1 3 7 10 14 


18 
17 


21 25 28 32 
20 24 27 31 


1173 1206 


1239 


1271 


1303 


1335 1367 


1399 


1430 


3 7 10 13 
3 7 10 12 


16 
16 


20 23 26 30 

19 22 25 29 


14G1 


1492 1523 


1553 


1584 


1 1614 


1644 1673 


1703 


1732 


3 6 9 12 
3 6 9 12 


15 
15 


18 21 24 28 
17 20 23 26 


15 
16 


1761 


1790 1818 1847 1875 

1 


1903 

1931 1959 1987 


2014 


3 6 9 11 
3 5 8 11 


14 
14 


. . 

17 20 23 26 
16 19 22 25 


•2011 


2068 


2095 


2122 


2148 1 

2175 1 2201 


2227 


2253 


2279 


3 5 8 11 
3 5 8 10 


14 
13 


16 19 22 24 
15 18 21 23 


17 
18 


2301 
2553 


2330 2355 


2380 


2405 1 2430 


1 2455 2480 


2504 


2529 


3 5 8 10 
2 5 7 10 


13 
12 


15 18 20 23 
15 17 19 22 


2577 I 2601 


2625 2648 f 

1 2672 


2695 


2718 


2742 


2765 


2 5 7 9 
2 5 7 9 


12 1 14 16 19 21 
11 14 16 18 21 


19 
20 


2788 


2810 


2833 


2856 


2878 1 

1 2900 


2923 


2945 


2967 


2989 


2 4 7 9 11 
2 4 6 8 11 


13 16 18 20 
13 15 17 19 


3010 


3032 


3054 


3075 


3096 1 3118 


3139 


3160 


3181 


3201 


2 4 6 8 11 1 13 15 17 19 


21 
22 
23 
2i 


3222 3243 3263 
3424 3444 3464 
3617 3636 ' 3655 
3802 3820 3838 


3284 
3483 
3674 
3856 


3304 J 3324 
3502 3522 
3692 I 3711 
3874 1 3892 


3345 
3541 
3729 
3909 


3365 
3560 
3747 
392-7 


3385 
3579 
3766 
3945 


3404 
3598 
3784 
3962 


2 4 6 8 , 10 12 14 16 18 
2 4 6 8 ; 10 12 14 15 17 
2 4 6 7 ! 9 ! 11 13 15 17 
2 4 5 7 ' 9 i 11 12 14 16 


25 


3979 


3997 4014 


4031 


4048 


4065 


4082 


4099 4116 


4133 


2 3 5 7 9 ! 10 12 14 15 


26 
27 
28 
29 

30 1 

31 
S2 
33 
34 

35 


4150 
4314 
4472 
4624 


4166 
4330 
4487 
4639 


4183 
4346 
4502 
4654 


4200 
4362 
4518 
4669 


1 
4216 
4378 
4533 
4683 


4232 
4393 
4548 
4698 


4249 
4409 
4564 
4713 


4265 
4425 
4579 

4728 


4281 
4440 
4594 
4742 


4298 
4456 
4609 
4757 


2 3 5 7 8 ! 10 1113 15 
2 3 5 6 8 1 9 11 13 14 
2 3 5 6 8 9 11 12 14 
13 4 6 7 9 10 12 13 


4771 


1 
4786 4800 


4814 


1 ^ 

4829 1 4843 


4857 


4871 


4886 


4900 


13 4 6 7 


9 10 11 13 


4914 

5051 
5185 
5315 


4928 j 4942 
5065 5079 
5198 1 5211 

5328 5340 

1 1 


4955 
5092 
5224 
5353 


4969 4983 
5105 5119 
5237 5250 
5366 5378 


4997 
5132 
5263 
5391 


5011 
5145 
5276 
5403 


5024 
5159 
5289 
5416 


5038 
5172 
5302 
5428 


13 4 6 17 
13 4 5 7 
13 4 5 6 
13 4 5 6 


8 10 11 12 
8 9 11 12 
8 9 10 12 
8 9 10 11 


5441 


5453 5465 


5478 


5490 


5502 


5514 


5527 5539 


5551 


1 
1 2 4 5 ' 6 


7 9 10 11 


36 55(33 

37 5U8:3 

38 5798 

39 5911 


5575 5587 
5694 5705 
5809 ; 5821 
5922 j 5933 


5599 

5717 
5832 
5944 


5611 
5729 
5843 
5955 


5623 
5740 
5855 
5966 


5635 

5752 
5866 
5977 


5647 5658 
5763 5775 
5877 5888 
5988 : 5999 


5670 
5786 
5899 
6010 


1 2 4 5; 6 
12 3 5 6 
12 3 5 6 
12 3 4 5 


7 8 10 11 
7 8 910 
7 8 910 

7 8 9 10 


40 1 

41 
42 
43 
44 


6021 


6031 6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


12 3 4 5 


6 8 9 10 


6128 
6232 
6335 
6435 


1 

6138 6149 
6243 6253 
6345 ' 6355 
6444 6454 j 


6160 
6263 
6365 
6464 


6170 

6274 
6375 
6474 


6180 
6284 
6385 
6484 


6191 
6294 
6395 
6493 


6201 
6304 
6405 ' 
6503 


6212 6222 
6314 6325 
6415 i 6425 
6513 j 6522 


i 

12 3 4 
12 3 4 
12 3 4 
12 3 4 


5 
5 
5 
5 


6 7 8 9 
G 7 8 9 
6 7 8 9 
6 7 8 9 


45 


6532 


6542 


1 
6551 €561 6571 


6580 


6590 


6599 1 6609 


6618 


12 3 4 5 

I 1 


6 7 8 9 


46 
47 
48 
49 


6628 
6721 
6812 
6902 


6637 
6730 
6821 
6911 


6646 6656 
6739 6749 
6830 6839 
6920 1 -6928 


6665 
6758 
6848 
6937 


6675 
6767 
6857 
6946 


6684 
6776 
6866 
6955 


1 

6693 6702 
6785 6794 
6875 1 6884 
6964 6972 


6712 
6803 
6893 
6981 


12 3 4 
12 3 4 
12 3 4 
12 3 4 


5 
5 
4 
4 


6 7 7 8 
5 6 7 8 
5 6 7 8 
5 6 7 8 


50 


6990 


6998 

j 


7007 7016 1 7024 


7033 


7042 


7050 


7059 


7067 


12 3 3 4 


5 6 7 8 



MATHEMATICAL TABLES 



597 



Logarithms 








1 


2 


3 


4 


3 


6 


7 


8 


i 9 


12 3 4 


5 


6 7 8 9 




51 
52 
63 

54 


7076 
7160 
7243 
7324 


7084 
7168 
7251 
7332 


7093 
7177 
7259 
7340 


7101 

7185 
7267 
7348 


7110 
7193 
7275 
7356 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 
7218 
7300 
7380 


7143 

7226 
7308 

7388 


i 
7152 
7235 
7316 
7396 


12 3 3 
12 2 3 
12 2 3 
12 2 3 


4 
4 
4 
4 


5 6 7 8 
56 7 7 
5 C 6 7 
5 6 6 7 




55 


7-404 


7412 


7419 


7427 


7435 1 7443 7451 


7459 


7466 


i 
7474 


12 2 3 


4 


5 5 6 7 




56 
57 
58 
59 


7482 
7559 
7G34 
7709 


7490 
7566 
7642 
7716 


7497 
7574 
7649 
7723 


7505 
7582 
7657 
7731 


7513 7520 
7589 7597 
7664 7672 
7738 7745 


7528 
7604 
7679 
7752 


7536 
7612 
7686 
7760 


7543 
7619 
7694 
7767 


7551 

7627 
7701 

7774 


12 2 3 
12 2 3 
112 3 
112 3 


4 
4 
4 

4 


5 5 7 
5 5 6 7 
4 5 6 7 
4 5 6 7 




60 


7782 


7789 


7796 


7803 


7810 1 7818 


7825 


7832 


7839 


7846 


112 3 


i 
4 


4 5 6 6 




61 
62 
63 
64 


7853 
7924 
7993 
8062 


7860 
7931 
8000 
8069 


7868 
7938 
8007 
8075 


7875 
7945 
8014 
8082 


7882 j 7889 
7952 j 7959 
8021 1 8028 
8089 j 8096 


7896 
7966 
8035 
8102 


' 7903 

' 7973 

8041 

i 8109 


7910 

7980 

8048 

■ 8116 


7917 
7987 
8055 
8122 


112 3 
112 3 
112 3 
112 3 


4 
3 
3 
3 


4 5 6 6 
4 5 6 6 
4 5 5 6 

4 5 5 6 




65 


8129 


8136 


8142 


8149 


8156 j 8162 


8169 


8176 


8182 


8189 


112 3 


3 


4 5 5 6 




63 
67 
68 
69 

70 


8195 
8261 
8325 

8388 


8202 
8267 
8331 
8395 


8209 
8274 
8338 
8401 


8215 
8280 
8344 
8407 


8222 8228 
8287 8293 
8351 8357 
8414 8420 


8235 
8299 
8363 
8426 


8241 
8306 
8370 
8432 


8248 
8312 
8376 
8439 


8254 
8319 
8382 
8445 


112 3 
112 3 
112 3 
112 2 


3 
3 
3 
3 


4 5 5 6 
4 5 5 6 
4 4 5 6 
4 4 5 6 




8451 


8457 


8463 


8470 


8476 1 8482 


8488 


1 8494 


8500 


J 8506 1 1 1 2 2 


3 4 4 5 6 




71 
72 
73 

7i 


8513 
8573 
8G33 

8692 


8519 
8579 
8639 
8698 


8525 
8585 
8645 
8704 


8531 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8543 
8603 
8663 
8722 


8549 
8609 
8669 

8727 


8555 
8615 
8675 
8733 


8561 
8621 
8681 
8739 


8567 
8627 
8686 
8745 


112 2 
112 2 
112 2 
112 2 


3 
3 
3 
3 


4 4 5 5 

4 4 5 5 
4 4 5 5 
4 4 5 5 




75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 1 1 1 2 2 


3 


3 4 5 5 




76 

77 
78 
79 


8808 
8865 
8921 
8976 


8814 

8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
8998 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 8854 
8904 8910 
8960 8965 
9015 9020 


8859 
8915 
8971 
9025 


112 2 
112 2 
112 2 
112 2 


3 
3 
3 
3 


3 4 5 5 
3 4 4 5 
3 4 4 5 
3 4 4 5 




80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 9074 


9079 112 2 


3 3 4 4 5 




81 
82 
83 
84 


9085 
9138 
9191 
9243 


9090 
9143 
9196 
9248 j 


9096 
9149 
9201 
9253 


9101 
9154 
9206 
9258 


9106 
9159 
9212 
9263 


9112 
9165 
9217 
9269 


9117 
9170 
9222 
9274 


9122 9128 
9175. 9180 
9227 9232 
9279 9284 


9133 
9186 
9238 
9289 


112 2 
112 2 
112 2 
112 2 


3 
3 

3 


3 4 4 5 
3 4 4 5 
3 4 4 5 
3 4 4 5 




85 


9294 


9299 i 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


112 2 


3 1 3 4 4 5 1 




86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9355 
9405 
9455 
9504 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9375 
9425 
9474 
9523 


9380 
9430 
9479 
9528 


9385 
9435 
9484 
9533 


9390 
9440 
9489 
9538 


112 2 
112 
112 
112 


3 
2 
2 

2 


3 4 4 5 
3 3 4 4 
3 S 4 4 
3 3 4 4 




90 


9542 


9547 


9552 


9557 '• 


9562 


9566 


9571 


9576 


9581 


9586 1 1 1 2 ; 


2 3 3 4 4 




91 
92 
93 
91 


9590 
9638 
9685 
9731 


9595 
9643 
9689 
9736 


9600 
9647 
9694 
9741 


9605 
9652 
9699 
9745 


9609 
9657 
9703 
9750 


9014 
9661 
9708 
9754 


9619 
9666 
9713 
9759 


9624 
9671 
9717 
9763 


9628 
9675 
9722 
9768 


9633 
9680 
9727 
9773 


112 
112 
112 
112 


1 
2 3 3 4 4 
2 3 3 4 4 
2 3 3 4 4 
2 3 3 4 4 




95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


112 


2 


3 3 4 4 




96 
97 
98 
99 


9823 
9868 
9912 
9956 


9827 
9872 
9917 
9961 


9832 
9877 
9921 
9965 


9836 , 
9881 
9926 
9969 


9841 
9886 
9930 
9974 


9845 
9890 
9934 
9978 


9850 i 
9894 
9939 
9983 


9854 
9899 
9943 
9987 


9859 
9903 
9948 
9991 


9863 
9908 
9952 
9996 


112 
112 

112 
112 


2 
2 
2 
2 


3 3 4 4 
3 3 4 4 
3 3 4 4 
3 3 3 4 





598 



MATHEMATICAL TABLES 

A>fTIL0GARITH3IS 






12 3 4 


•5 


6 7 8 9 


12 3 4 


5 6 7 8 9 




•00 vy»-> 


l<yy2 lOOb 10i:'7 K»y> 


i':'i2 


1"14 1016 1019 1021 


M n 1 1 


1 1 ■. ■ ■- 




•01 1V23 
•02 l':»47 
•03 1072 
•04 lt»'. 


10-26 1(K8 1 1030 1033 
1050 105-2 lO-M 1057 
1074 1076 1079 lOSl 
1099 ll'>2 ll'H 111' 7 


1035 
1059 
1084 

iit:'9. 


1 1 

1038 1 1(M0 1042 1045 
1062 1064 1067 1069 
lt)86 1089 1091 1094 

1112 1114 1117 1119 


11 
11 
11 

111 


112 2 2 
112 2 2 
112 2 2 

1 2 2 2 2 




■05 1122 


1125 1 U27 ll:5r. 1132 


11:55 


ll:>^ 1140 1143 1146 


"111 


1 2 2 2 2 




•06 11 tS 
•07 1175 

•08 la^yi 

•09 12:J») 


1151 U-53 1156 1159 
117S 11S«:» ll'^S 11S6 
1205 12«>? 1211 1213 
1233 1238 1239 1242 


1161 
1189 
1216 
1245 


1164 1167 1169 1172 
1191 IIW 1197 1199 
1219 12-22 12-25 1227 
1247 1250 1253 1256 


111 
111 
111 
111 


12 2 2 2 
12 2 2 2 
12 2 2 3 
1 2 2 2 3 




•10 1251 


1262 j 12>>5 12>'^^ 1271 


1274 


127t' li79 1282 12--'> 


l' 1 1 1 


12 2 2 3 




•11 12^8 
•12 131S 
•13 1:549 
■14 i:>-'-^ 


1291 12W 1297 WX) 
1321 1324 1327 1330 
1352 1355 135S 1361 
13S4 13S7 1390 1393 


1.3i)3 
1334 
136v5 
1396 


lZty> 1309 1312 1315 
1:537 1340 1343 1346 
i:56^ 13n 1374 1377 
14'» 1403 1406 I4i>9 


111 

f) 1 1 1 

111 

'"' 1 1 1 


2 2 2 2 3 
2 2 2 2 3 
2 2 2 3 3 




■15 1 1413 1416 1419 li22 142^' 


1429 1 1432 1435 1439 1U2 


'^ 1 1 1 


2 2 2 3 3 




•16 1445 
■17 1479 
•18 1-514 
•19 1-549 


1449 1452 ' 1455 1459 
14S3 1486 1489 1493 
1517 1521 1524 152S 
155J 1556 1560 1563 


1462 
1496 
1-531 
1-567 


1466 1469 1472 1476 
1500 1503 1507 1510 
1-535 1538 1542 1545 
1570 1574 1578 15^1 


111 
111 
111 
111 


2 2 2 3 3 
2 2 2 3 3 
2 2 2 3 3 
2 2 3 3 3 




■20 1">5 


15S9 


1592 1596 1600 


i^^.s 


1>V'7 l.Ul 1614 l.;l> 


i. 1 1 1 


2 2 3 3 3 




21 1'22 
•22 1»'.'V' 
•23 1»;'.'> 
•24 173> 


16-26 
16»53 
17t>2 

1742 


1629 1633 1637 
1667 1671 1675 
1706 inO 1714 

1746 17-50 I7.54 


i>:v4i 

1679 
1718 

17-5S 


1>U4 1648 1652 lt'.56 
1683 16'<7 1690 1694 
1722 1726 1730 1734 
17H2 1766 I77t> 1774 


» 1 1 2 
112 
112 
112 


2 2 3 3 3 
2 2 3 3 3 
2 2 3 3 4 

2 2 3 3 4 




■25 177- 


17-1 17-'> 1791 1795 


17'j9 


1.^:13 lNi7 I'-ll l>lr; 


i< 1 1 2 


2 2 3 3 4 




•26 1-2'. 
•27 1-2 
•23 r.v»5 
29 1950 


1S24 1S2S 
1S»56 i 1871 
1910 ' 1914 
1954 1959 


1832 1S37 
1875 IS 79 
1919 19-23 
1963 196.^ 


1>4] 
1S.<4 
19"2^ 


l-<45 1,S49 1JS54 18-58 
1888 1892 1S97 19<Jl 
19:52 1936 1941 1945 
1^77 19S2 1986 1991 


112 
112 
112 

n 1 1 2 


2 3 3 3 4 
2 3 3 3 4 

2 3 3 4 4 

2 3 3 4 4 




•30 1995 


2»XjO 2004 


2009 , 2014 


2"1- 


2i>23 ' 2<>28 2032 2ii37 


1:' 112 


2 3 3 4 4 




■31 21)42 
■32 2089 
•33 213-S 
■34 21SS 


2046 2(^1 
2«)94 2099 
2143 ' 2148 
2193 2198 


2056 1 2061 
2104 2109 
21>3 2158 
2-203 2208 


2«>55 
2113 
216« 
2213 


2070 2075 2060 20W 
2118 -2123 2128 2133 
2168 2173 2178 2183 
2218 2-2-23 2-2-2S 2234 


112 
112 
112 
112 2 


2 3 3 4 4 
2 3 3 4 4 

2 3 3 4 4 

3 3 4 4 5 




•35 2239 

•36 -2291 
37 2Ui 
•38 2399 
■39 24-55 


2244 


1 
2249 2254 2259 


■1-^,:', 


2270 -2275 _ 22.^.' 22>^; 


112 2 


3 3 14 5 




2296 
2350 
2404 

2460 


2o<:>l 2307 2312 
2355 2360 2366 
2410 -2415 2421 

2466 2472 2477 


2317 
2371 
2427 
24>3 


2323 2328 ! 2333 "2339 
2377 2382 2388 2393 
2432 24.38 -2443 -2449 

24*^9 2495 -Z'/*^ 2-V>6 


112 2 
112 2 
112 2 
112 2 


3 3 4 4 5 
3,3445 
3 3 4 4 5 

3 3 4 5 5 


• 


■40 2512 


2.n- 2-523 2529 2535 


2->4l 


2-547 2-5-53 2 -55 'J 2-'>'-4 


1 1 2 2 


344 5 5 




41 2570 

42 -2630 

43 2>?92 

44 27->4 

•45 2S1> 


2576 2582 2588 2594 
2636 2642 2649 2655 
2698 2704 2710 2716 
2761 2767 2773 27SO 


26ij"» 
2661 
2723 


2606 2*512 2618 2624 
2667 -2673 2679 2685 
2729 2735 2742 2748 
2793 2799 2805 -2812 


112 2 
112 2 

112 3 

112 3 


3 4455 
3 4 4 5 6 
3 4 4 5 6 

3 4 4 5 6 




2>25 2S31 2838 2844 


2>51 


28-58 2864 2>71 2877 


112 3 


3 4 5 5 6 




•46 2SS4 

■47 -yyc'i 

■48 :;.'LV 
•49 :.■■>: 


•^91 2897 2904 2911 
2958 2965 2972 2979 
3»>27 3034 3041 3048 
3^97 3105 3112 3119 


2917 
2985 
3055 

31-26 


2924 2931 2938 29+4 
2999 2992 3006 3013 
30»)2 3069 3076 3083 
31-^3 3141 3148 31-55 


112 3 
112 3 
112 3 
112 3 


3 4 5 5 6 

3 4 5 5 6 

4 4 5 6 6 
4 4 5 6 8 





MATHEMATICAL TABLES 
Ajsttilogarithms 



599 










1 ; 2 3 

i i 


4 


5 


1 

6 


7 


8 


9 


1 


2 


34 


5 


6 7 8 9 




•50 


3162 


1 1 

3170 1 3177 3184 


3192 


3199 


i 

3206 j 


3-314 


3221 


3228 


1 


1 


2 


3 


4 


4 5 6 7 




•51 
•52 
•53 
•54 


3236 
3311 
3388 
3467 


3243 ' 3251 3258 
3319 i 3327 3334 
3396 ' 3404 3412 
3475 , 3483 3491 


3266 
3342 
34-20 
3499 


3273 
3350 
3428 
3508 


3281 
3357 
3436 
3516 


3289 
3365 
3443 
3524 


3296 
3373 
3451 
3532 


3304 
3381 
3459 
3540 


1 

1 
1 
1 


2 

2 
2 
2 


2 
2 

2 
2 


3 
3 
3 
3 


4 

4 ■ 

4 

4 


5 5 6 7 
5 5 6 7 
5 6 6 7 
5 6 6 7 




•55 


3548 


3556 1 3565 


3573 


3581 


3589 


3597 


3606 


3614 


3622 


1 


2 


2 


3 


4 

1 


5 6 7 7 




•56 
•57 
•58 
59 


3631 
3715 

3802 
3890 


3639 3648 
3724 3733 
3811 3819 
3899 3908 


3656 
3741 

3828 
3917 


3664 
3750 
3837 
3926 


3673 
3758 
3846 
3936 


3681 
3767 
3855 
3945 


3690 
3776 
3864 
3954 


3698 
3784 
3873 
3963 


3707 
3793 
3882 
3972 


1 
1 
1 
1 


2 
2 
2 
2 


3 
3 
3 
3 


3 
3 
4 
4 


4 

4 . 
4 
5 


5 6 7 8 
5 6 7 8 
5 6 7 8 
5 6 7 8 




•60 


3981 


3990 3999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


1 


2 


3 


4 


5 


6 6 7 8 




•61 
•62 
•63 
•64 


4074 
4169 
4266 
4365 


4083 
4178 
4276 
4375 


4093 
4188 
4285 
4385 


4102 
4198 
4295 
4395 


4111 
4207 
4305 
4406 


4121 
4217 
4315 
4416 


4130 
4227 
4325 
4426 


4140 
4236 
4335 
4436 


4150 
4246 
4345 
4446 


4159 
4256 
4355 
4457 


1 
1 
1 

1 


2 

2 
2 
2 


3 
3 
3 
3 


4 
4 
4 
4 


5 
5 

5 
5 


6 7 8 9 
6 7 8 9 
6 7 8 9 
6 7 8 9 




•65 


4467 


4477 


4487 


4498 


4508 


4519 


4529 


4539 


4550 


4560 1 1 


2 


3 


4 


5 


6 7 8 9 




•66 
•67 
•68 
•69 


4571 
4677 
4786 

4898 


4581 
4688 
4797 
4909 


4592 
4699 
4808 
4920 


4603 
4710 
4819 
4932 


4613 
4721 
4831 
4943 


4624 
4732 
4842 
4955 


4634 
4742 
4853 
4966 


4645 
4753 
4864 
4977 


4656 
4764 
4875 
4989 


4667 
4775 

4887 
5000 


1 
1 
1 
1 


2 
2 
2 
2 


3 
3 
3 
3 


4 
4 
4 
5 


5 
5 
6 
6 


6 7 910 

7 8 9 10 
7 8 910 
7 8 910 




•70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


5093 


5105 


5117 


1 


2 


4 


5 


6 


7 8 911 




•71 

•72 
•73 

•74 


5129 
5248 
5370 
5495 


5140 5152 
5260 5272 
5383 5395 
5508 5521 


5164 
5284 
5408 
5534 


5176 
5297 
5420 
5546 


5188 
5309 
5433 
5559 


5200 
5321 
5445 
5572 


5212 
5333 
5458 

5585 

t 


5224 
5346 
5470 
.5598 


5236 
5358 
5483 
5610 


1 
1 
1 

1 


2 
2 
3 
3 


4 
4 
4 
4 


5 
5 
5 
5 


6 
6 
6 
6 


7 8 10 11 

7 9 10 11 
8. 9 10 11 

8 9 10 12 




•75 


5623 


5636 5649 


5662 


5675 


5689 


5702 


j 5715 


5728 


5741 


1 


3 


4 


5 


7 


8 9 1012 




•76 

•77 
■78 
•79 


5754 

5888 
6026 
6166 


5768 
5902 
6039 
6180 


5 781 
5916 
6053 
6194 


5794 
5929 
6067 
6209 


5808 
5943 
6081 
6223 


5821 
5957 
6095 
6237 


5834 
5970 
6109 
6252 


5848 

5984 

6124 

i 6266 


5861 
5998 
6138 
6281 


5875 
6012 
6152 
6295 


1 
1 
1 

1 


3 
3 
3 
3 


4 
4 
4 
4 


5 
5 
6 
6 


7 

7 
7 
7 


8 9 11 12 
8 10 11 12 

8 10 11 13 

9 10 11 13 




•80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


6412 


6427 


6442 


1 


3 


4 


6 


7 


9 10 12 13 




•81 
•82 
•83 
•84 


6457 
6607 
6761 
6918 


6471 
6622 
6776 
6934 


6486 
6637 
6792 
6950 


6501 

6653 

6808 

' 6966 


6516 
6668 
6823 
6982 


6531 
6683 
6839 
6998 


6546 
6699 
6855 
7015 


! 6561 

i 6714 

; 6871 

7031 


6577 
6730 
6887 
7047 


6592 
6745 
6902 
7063 


2 
2 
2 
2 


3 
3 
3 
3 


5 
5 
5 
5 


6 
6 
6 
6 


8 
8 
8 
8 


9 11 12 14 

9 11 12 14 

9 11 13 14 

10 11 13 15 




85 


7079 


7096 


7112 7129 


, 7145 


7161 


7178 


7194 


7211 


7228 


2 


3 


5 


7 


8 


10 12 13 15 




-88 
•87 
•83 
•89 


7244 
7413 
7586 
7762 


7261 
7430 
7603 

7780 


7278 7295 
7447 1 7464 
7621 7638 
7798 ; 7816 


7311 

7482 
7656 
7834 


7328 
7499 
7674 

7852 


7345 
7516 
7691 

7870 


: 7362 

I 7534 

7709 

i 7889 


7379 
7551 
7727 
7907 


1 7396 

. 7568 

7745 

7925 


2 
2 
2 
2 


3 
3 
4 
4 


5 
5 
5 
5 


7 
7 
7 
7 


8 
9 
9 
9 


10 12 13 15 

10 12 14 16 

11 12 14 16 
11 13 14 16 




•90 


7943 


7962 


7980 i 7998 


, 8017 


8035 


8054 


1 
8072 


8091 


8110 


2 


4 


6 


7 


9 


11 13 15 17 




•91 
•92 
•93 
•94 


8128 
8318 
8511 
8710 


8147 
8337 
8531 
8730 


8166 
8356 
8551 
8750 


8185 
8375 
8570 
8770 


8204 
8395 
8590 
8790 


8222 
8414 
8610 
8810 


8241 
8433 
8630 
8831 


1 
8260 
8453 
8650 

8851 


8279 
8472 
8670 

8872 


1 8299 

' 8492 

8690 

1 8892 


2 
2 
2 
2 


4 
4 
4 
4 


6 
6 
6 
6 


8 
8 
8 
8 


9 
10 
10 
10 


11 13 15 17 

12 14 15 17 
12 14 16 18 
12 14 16 18 




•95 


8913 


8933 


8954 


8974 


.8995 


9016 


9036 


9057 


9078 


9099 1 2 


4 


6 


8 


10 


12 15 17 19 




•96 
•97 
•98 


9120 
9333 
9550 


9141 
9354 
9572 


9162 
9376 
9594 


9183 
9397 
9616 


! 9204 
9419 
9638 


9226 
9441 
9661 


9247 
9462 
9683 


1 

i 9268 
1 9484 
' 9705 


9290 
9506 
9727 


9311 
9528 
9750 


2 
2 
2 


4 
4 
4 


6 

7 
7 


8 
9 
9 


11 
11 
11 


13 15 17 19 
13 15 17 20 
13 16 18 20 



INDEX 



Abrupt change of section, 56, 90 
Adams's experiments on marble, 

42 
Adhesion between concrete and 

steel, 80 
Alignment charts for springs, 342 
Aluminium, 62 
Andrews -Pearson formula for 

curved beams, 545 
Areas — 

mathematical determination, 
161 

Parmontier's rule, 164 

Simpson's rule, 164 

sum curves, 162 

table of various sections, 185 
Arnold's testing machine, 399 
Autographic recorders, 379 
Avery's reverse torsion machine, 

385 

Bach on plates and slabs, 488-509 

Bairstow, 59, 88 

Baker, B. — 

abrupt change of section ex- 
periments, 56 
repetitions of stress, 86 

Basquin, 307 

Bauschinger on strength of stone, 
72 

Beam factor, 197 

Beams. See Bending moments; 
deflections ; inclined beams ; 
shear; stresses, bending. 

Bending moments — 
cantilevers, 122-127 
continuous beams, 430-458 
fixed beams, 414-438, 458 
simply supported beams, 127- 
160 

Bernoulli's assumption, 194 

Bolts, coupling, 311 

Brinell hardness test, 402 



Buckton testing machines, 364, 
368, 381 

Built-in beams, 416-438, 458 

Bouton experiments on compres- 
sion, 47 

Bracing of columns, 294 

Brass, 62, 83 

Brittleness, 41 

Bronze, 62, 83 

Buchanan on columns, 287 

Buckling factor, 279 

Bulk modulus, 7 

Cantilevers, 122-127 

Cast iron, 49, 209, 305 

Centroid, 165 

Chain links, 546-551 

Cleat, stresses in rivets, 315 

Coker — 

thermal and optical testing, 

409, 411 
torsion testing, 388 

Collapsing pressure for pipes, 
118 

Columns — 

centrally loaded, 279-301 
eccentrically loaded, 301-310 

Compressive strength, see various 
materials. 

Concrete — 

compressive strength, 69-77, 82 
reinforced, 215-235, 292, 479 
shear strength, 78 
tensile strength, 77 

Considere, 42 

Continuous beams, 438-458 

Critical speed of shafts, 562-569 

Curvature of beams, 250 

Curved beams — 

Andrews- Pearson formula, 545 
correction coefficients, 543 
general conditions of strain, 529 
Resal's construction, 538 



601 



602 



INDEX 



Ctirved beams {continued) — 
rings and chain links, 546 
Winkler's formula, 533 

Cylinders, see Pipes. 

Darwin's extensometer, 376 
Deflections — 

beams, simply supported and 
cantilever, 248-278 

fixed beams, 419, 420, 460 

shear, due to, 483 
Diagonal square beam section, 

210 
Disks, rotating, 552-562 
Distribution of shear stress in 

beams, 470-487 
Dixon and Hummel testing 

machine, 368 
Drums, rotating, 552 
Ductility, 41 
Dunkerley on whirling of shafts, 

568 
Dynamic stress and strain, 33 
Dyson on internal friction, 47 

Eden on stress repetition, 89 
Elastic bodies, 1 
Elastic moduli, 7 

relation between, 8 
Ellipse — 

of inertia, 169 

of stress, 15 
Elongation and gauge length, 54 
Encastre beams, 416^38, 458 
Euler's column formula, 280, 306 
Ewing's extensometer, 374 

Failure, cause of, 42 
Fairbaim, 84, 119 
Fatigue of metals, 89 
Fidler column formula, 289 
Filon on optical testing, 415 
Fitchett, F., on springs, 343 
Fixed beams, 416-438, 458 
Flitched beams, 203 
Foster on stress rej)etition, 89 

Gadd on cement tests, 408 
Glass, 83 

Goodenough on chain links, 551 
Goodman extensometer, 371 
Gordon column formula, 288 
Grashof on plates and slabs, 494 



Greenwood and Batley testing 

machine, 367 
Grips for test-pieces, 370 
Guest theory of stress, 44, 328 

Hardness, 41, 402 

Hartnell, W., on springs, 341 

Heterogeneous sections — 

direct stress, 38 

geometrical properties, 183 
Hooke's Law, 2 

Horse- power of shafting, 323, 335 
Hysteresis, mechanical, 59, 89 

Illinois experiments, 119, 297, 551 
Impact, strain and stress due to, 35 
Impact testing, 400 
Inclined beams, 155-160 
Inertia, moment of, 169-191 
Internal friction in materials, 45 
Izod — 

exjjeriments on shear, 64, 67 
impact testing machine, 401 

Johnson — 

column formula, 288 
eccentric loading, 309 

Kennedy's extensometer, 372 
Keyways in shafting, 334 

Lame's theory for thick pipes, 510 

Lilly- 

column formula, 290 
torsion testing machine, 386 

Live loads, 100 

Liider's Lines, 54 

Macklow-Smith torsion meter, 392 
Malleability, 41 
Malleable cast iron, 51 
Modulus — 

elastic, 7 

section, 197 
Mohr, 181, 252 
Moment of inertia, 167-191 
Moment of resistance, 196, 213, 222 
Momental ellipse, 167 
Moore, 297, 334, 551 
Morley, 3 

on columns, 310 

on curved beams, 545 

on slabs, 494 
Muir on overstrain, !59 



INDEX 



603 



Navier internal friction theory, 45 
Non-circular shafts, 333 

Oblique loading on beams, 241 
Overstrain, 57 

Permanent set, 1 

Perry, 70, 192 

Pipes and cylinders — 

collapse of, 118 

initial pressure in, 524 

Lame's theory, 510 

shear stresses in, 51 6 

thick, 510-538 

thin, 115 
Piston rings, 355 
Plastic bodies, 1 
Plates — 

Bach theory, 488^93, 499-509 

circular, 488 

Grashof and Rankine, 494 

oval, 492 

square and rectangular, 494 
Poisson's ratio, 3 
Polar moment of inertia, 173 
Portland cement, strength of, 68- 

79, 404 
Principal stresses, 13 

Quality factor, 62 

Radius of gyration,' 169-191 
Rankine — 

column formula, 285 

combined stress theory, 43, 
320 

slab formula, 494 

stress lines, 213 
Reinforced concrete — 

beams, 215-235 

columns, 292 

shear in beams, 479 
Repetition of stress, 84 
Resal's construction, 538 
Resilience — 

definition, 33 

in bending, 272 

in torsion, 331 

summary for various springs, 
363 
Rigidity modulus, 7 
Rings, 546 
Riveted joints, 102-115 



Rotating drums and disks, 550- 

562 
Rotating shafts, 562-569 

St. Venant, 44, 328, 333 
Sankey bending machine, 396 
Scoble on optical testing, 404 
Secondroid, 168 
Section modulus, 197 
Shaft coupling, stresses in, 311 
Shafting, stresses in, 316-335 
Shear — 

diagrams for cantilevers, 122- 
127 

diagrams for simply supported 
beams, 127-160 

diagrams, steps in, 144 

strain and stress, 3 

stress and strain equivalent to 
complex stresses, 19, 30 

stress in beams, 470 
Shrinkage stresses in pipes, 525 
Slabs, see Plates. 
Slate, 83 

Slenderness ratio, 279 
Smith, C. A. M., 48, 517 
Smith, J. H., repetition testing 

machine, 392 
Smith, R. H., construction for 

combined stresses, 23 
Springs — 

closed-coiled helical, 337, 362 

leaf on plate, 351 

open-coiled helical, 346 

piston rings, 355 

plane spiral, 360 

summary, 363 

time of vibration, 337 
Stanchions, see Columns. 
Stewart on collapse of pipes, 

119 
Stone, compressive strength, 68, 

82 
Straight line column formula, 287 
Strain — 

definition and kinds, 1-3 

in different directions, 12 

maximum strain equivalent to 
combined strains, 25 

transverse, 3 
Stress — 

bending, 192-247 

cause of failure under, 42 



604 



INDEX 



Stress (continued) — ■ 

combined bending and direct, 
235-240, 328 

complex, 13 

definitions and kinds, 1-3 

dynamic, 33 

ellipse of, 15 

impact, 35 

principal, 13, 19 

repetition of, 84 

shear, 3, 212 

shear in beams, 470 

temperature, 37 

working, 93-100 
Stress-strain diagrams, 5, 49, 53, 

60, 62, 65 
Struts, see Columns. 
Sum curve, 162 
Superposition, principle of, 244 

T beams, reinforced concrete, 229 
Temperature stresses, 37 

effect on strength, 63 
Tensile strength, real and appar- 
ent, 52 
of materials, see various ma- 
terials. 
Testing — 

calibration of machines, 369 
cement and concrete, 404 
extensometers, 371-380 
grips and forms of test- pieces, 

370 
impact, hardness and ductility, 

396 
machines (general), 364-369 



Testing (contimied)— 

optical, 411 

thermal, 409 

torsion machines, 382-393 
Theorem of three moments, 445 
Thick pipes, see Pipes. 
Thurston testing machine, 383- 

386 
Timber, 65, 82 
Torsion, 311-335 
Turbine shaft, settling down, 564 
Twisting, 311-335 

Unit section modulus, 197 
Unital strain, 3 
Unwin — 

elongation formula, 55 

extensometer, 379 

on piston rings, 359 

on stone cubes, 70 

Waist in tensile fracture, 54, 56 
Werder testing machine, 366 
Whirling of shafts, 562-569 
Wicksteed-Buckton testing- ma- 
chines, 364, 368, 381 
Winkler's formula for curved 

beams, 533 
Wohjer's experiments, 84 
Woo J son experiments on concrete, 

43 
Working stresses, 93-99 
Wrought iron, 82 

Yield point, 4 
Young's Modulus, 7 



Printed ix Gkeat Bkitain- bv Richaud Clav & Soss, Limited. 

IIRLXSWICK. ST., STAMFOKD ST., S.E., AND hlNGAV, Sl'FFOLE. 



^^ -^A 



0' 



